UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-30T05:11:07Z

JonathanRothwell: /* Richter Magnitude Scale */

==Logarithmic Scale==
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| LOGARITHMIC SCALE
|-
|style="background: #FBBBB9; padding:12px;"|
A '''Logarithmic Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithm scale.
|}
==Richter Magnitude Scale==

[[File:Rc.png]]

Invented by Charles Richter in 1935, the Richter Scale or the
'''Local Magnitude Scale (ML)''' measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake.

The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is '''tenfold.''' This means that each number increase on the Richter scale indicates an intensity ten times stronger. The range of the Richter scale is between 0 and 10. An earthquake can measure above 10.0, which is then called an epic earthquake.

An earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is '''10 x 10 = 100''' times stronger than an earthquake of magnitude 5. An earthquake of magnitude 8 is ''' 10 x 10 x 10 = 1000''' times stronger than an earthquake of magnitude 5. In terms of energy, a whole number change approximately corresponds to 31 times more or less energy being released <ref>USGS, 2009
http://earthquake.usgs.gov/learn/topics/richter.php
</ref>.

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Comparing Earthquakes'''
|-
|style="background: #ADDFFF; padding:12px;"|

How does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
*The increase in magnitude is 5-2 or 3.

*<math>10^3 = 1000</math>; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

How does a magnitude '''7''' quake compare to a magnitude '''1''' earthquake?
*The increase in magnitude is 7-1 or 6.
*<math>10^6=1 000 000</math> thus the ground movement of a magnitude 7 quake is '''one million times greater''' than a magnitude 1 quake.

'''
As illustrated in the examples above, there is are large differences between the impact of earthquakes because of the Richter scale being a logarithmic scale.'''

|}
[[File:Richterscale.png]]
----

The Richter magnitude of an earthquake is determined from the
logarithm of the amplitude of waves recorded by seismographs. </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref> The original formula is: [[File:Log.png]]

Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by '''"R= log[ I ÷ I0]"'''
*I0 is the "threshold quake", or movement that can barely be detected and I, the intensity, is given in terms of multiples of that threshold intensity.
*Suppose a seismograph measures that the I= 989I0.
**To determine the magnitude of the earthquake, the intensity to a Richter rating by evaluating the Richter function at '''I= 98910:R = log[I ÷ I0 ]=log[ 98910 ÷ I0] =log[989]=2.99519629'''... or about 3 on the Richter Scale, which turns out to be a minor earthquake.
----

{{#ev:youtube | tcmrlR2XMNM| 400}}

----

''' References:''' <references/>

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-30T05:04:24Z

JonathanRothwell: /* Richter Magnitude Scale */

==Logarithmic Scale==
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| LOGARITHMIC SCALE
|-
|style="background: #FBBBB9; padding:12px;"|
A '''Logarithmic Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithm scale.
|}
==Richter Magnitude Scale==

[[File:Rc.png]]

Invented by Charles Richter in 1935, the Richter Scale or the
'''Local Magnitude Scale (ML)''' measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake.

The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is '''tenfold.''' This means that each number increase on the Richter scale indicates an intensity ten times stronger. The range of the Richter scale is between 0 and 10. An earthquake can measure above 10.0, which is then called an epic earthquake.

An earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is '''10 x 10 = 100''' times stronger than an earthquake of magnitude 5. An earthquake of magnitude 8 is ''' 10 x 10 x 10 = 1000''' times stronger than an earthquake of magnitude 5. In terms of energy, a whole number change approximately corresponds to 31 times more or less energy being released.

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Comparing Earthquakes'''
|-
|style="background: #ADDFFF; padding:12px;"|

How does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
*The increase in magnitude is 5-2 or 3.

*<math>10^3 = 1000</math>; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

How does a magnitude '''7''' quake compare to a magnitude '''1''' earthquake?
*The increase in magnitude is 7-1 or 6.
*<math>10^6=1 000 000</math> thus the ground movement of a magnitude 7 quake is '''one million times greater''' than a magnitude 1 quake.

'''
As illustrated in the examples above, there is are large differences between the impact of earthquakes because of the Richter scale being a logarithmic scale.'''

|}
[[File:Richterscale.png]]
----

The Richter magnitude of an earthquake is determined from the
logarithm of the amplitude of waves recorded by seismographs. </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref> The original formula is: [[File:Log.png]]

Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by '''"R= log[ I ÷ I0]"'''
*I0 is the "threshold quake", or movement that can barely be detected and I, the intensity, is given in terms of multiples of that threshold intensity.
*Suppose a seismograph measures that the I= 989I0.
**To determine the magnitude of the earthquake, the intensity to a Richter rating by evaluating the Richter function at '''I= 98910:R = log[I ÷ I0 ]=log[ 98910 ÷ I0] =log[989]=2.99519629'''... or about 3 on the Richter Scale, which turns out to be a minor earthquake.
----

{{#ev:youtube | tcmrlR2XMNM| 400}}

----

''' References:''' <references/>

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:22:05Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is <math>y=|x|</math>, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like <math>-\frac{1}{2}|x+3|-1?</math>

*When you have a function in the form <math>y = |x + h|</math> the graph will move h units to the left.
*When you have a function in the form <math>y = |x - h|</math> the graph will move h units to the right.
*When you have a function in the form <math>y = |x| + k</math> the graph will move up k units.
*When you have a function in the form <math>y = |x| - k</math> the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what is inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left <math>y=|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the <math>-\frac{1}{2}</math> coefficient <math>y= -\frac{1}{2}|x+3|</math>

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*<math>-|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*<math>-\frac{1}{2}|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down <math>-\frac{1}{2}|x+3|-1</math>
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If <math>x > = 0</math> then <math>| x | = x</math> and if <math>x < 0</math> then <math>| x | = -x</math>.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

<math>- 2 + 10 = 8</math> is positive and according to the definition above <math>| -2 + 10 |</math> can be simplified as follows:

<math>| -2 + 10 | = | 8 | = 8</math>

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*<math>| 2x + 1 |</math>

*<math>| x + 3| , if x < -3</math>

*<math>| -x + 2 | , if x > 2</math>

'''Solution to Example 2'''

<math>2x + 1</math> is always positive and according to the definition above <math>| 2x + 1 |</math> can be simplified as follows:

<math>| 2x + 1 | = 2x + 1</math>

If <math>x < -3</math>, then <math>x + 3 < 0</math>. According to the definition above <math>| x + 3 |</math> can be simplified as follows:

<math>| x + 3 | = -(x + 3) = -x - 3</math>

If <math>x > 2</math> then <math>x - 2 > 0</math> and <math>-x + 2 < 0</math>. According to the definition above <math>| -x + 2 |</math> can be simplified as follows:

|-x + 2 | = - ( - x + 2 )= x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:21:07Z

JonathanRothwell: /* What are they? */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is <math>y=|x|</math>, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like <math>-\frac{1}{2}|x+3|-1?</math>

*When you have a function in the form <math>y = |x + h|</math> the graph will move h units to the left.
*When you have a function in the form <math>y = |x - h|</math> the graph will move h units to the right.
*When you have a function in the form <math>y = |x| + k</math> the graph will move up k units.
*When you have a function in the form <math>y = |x| - k</math> the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what is inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left <math>y=|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the <math>-\frac{1}{2}</math> coefficient <math>y= -\frac{1}{2}|x+3|</math>

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*<math>-|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*<math>-\frac{1}{2}|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down <math>-\frac{1}{2}|x+3|-1</math>
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If <math>x > = 0</math> then <math>| x | = x</math> and if <math>x < 0</math> then <math>| x | = -x</math>.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

<math>- 2 + 10 = 8</math> is positive and according to the definition above <math>| -2 + 10 |</math> can be simplified as follows:

<math>| -2 + 10 | = | 8 | = 8</math>

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*<math>| 2x + 1 |</math>

*<math>| x + 3| , if x < -3</math>

*<math>| -x + 2 | , if x > 2</math>

'''Solution to Example 2'''

<math>2x + 1</math> is always positive and according to the definition above <math>| 2x + 1 |</math> can be simplified as follows:

<math>| 2x + 1 | = 22 + 1</math>

If <math>x < -3</math>, then <math>x + 3 < 0</math>. According to the definition above <math>| x + 3 |</math> can be simplified as follows:

<math>| x + 3 | = -(x + 3) = -x - 3</math>

If <math>x > 2</math> then <math>x - 2 > 0</math> and <math>-x + 2 < 0</math>. According to the definition above <math>| -x + 2 |</math> can be simplified as follows:

|-x + 2 | = - ( - x + 2 )= x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:20:21Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is <math>y=|x|</math>, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like <math>-\frac{1}{2}|x+3|-1?</math>

*When you have a function in the form <math>y = |x + h|</math> the graph will move h units to the left.
*When you have a function in the form <math>y = |x - h|</math> the graph will move h units to the right.
*When you have a function in the form <math>y = |x| + k</math> the graph will move up k units.
*When you have a function in the form <math>y = |x| - k</math> the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what is inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left <math>y=|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the <math>-\frac{1}{2}</math> coefficient <math>y= -\frac{1}{2}|x+3|</math>

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*<math>-|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*<math>-\frac{1}{2}|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down <math>-\frac{1}{2}|x+3|-1</math>
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If <math>x > = 0</math> then <math>| x | = x</math> and if <math>x < 0</math> then <math>| x | = -x</math>.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

<math>- 2 + 10 = 8</math> is positive and according to the definition above <math>| -2 + 10 |</math> can be simplified as follows:

<math>| -2 + 10 | = | 8 | = 8</math>

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*<math>| 2x + 1 |</math>

*<math>| x + 3| , if x < -3</math>

*<math>| -x + 2 | , if x > 2</math>

'''Solution to Example 2'''

<math>2x + 1</math> is always positive and according to the definition above <math>| 2x + 1 |</math> can be simplified as follows:

<math>| 2x + 1 | = 22 + 1</math>

If <math>x < -3</math>, then <math>x + 3 < 0</math>. According to the definition above <math>| x + 3 |</math> can be simplified as follows:

<math>| x + 3 | = -(x + 3) = -x - 3</math>

If <math>x > 2</math> then <math>x - 2 > 0</math> and <math>-x + 2 < 0</math>. According to the definition above <math>| -x + 2 |</math> can be simplified as follows:

|-x + 2 | = - ( - x + 2 )= x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:14:49Z

JonathanRothwell: /* Graphing Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is <math>y=|x|</math>, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like <math>-\frac{1}{2}|x+3|-1?</math>

*When you have a function in the form <math>y = |x + h|</math> the graph will move h units to the left.
*When you have a function in the form <math>y = |x - h|</math> the graph will move h units to the right.
*When you have a function in the form <math>y = |x| + k</math> the graph will move up k units.
*When you have a function in the form <math>y = |x| - k</math> the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what is inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left <math>y=|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the <math>-\frac{1}{2}</math> coefficient <math>y= -\frac{1}{2}|x+3|</math>

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*<math>-|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*<math>-\frac{1}{2}|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down <math>-\frac{1}{2}|x+3|-1</math>
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:12:17Z

JonathanRothwell: /* Graphing Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like <math>-\frac{1}{2}|x+3|-1?</math>

*When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what is inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the <math>-\frac{1}{2}</math> coefficient <math>y= -\frac{1}{2}|x+3|</math>

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*<math>-\frac{1}{2}|x+3|</math>
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down <math>-\frac{1}{2}|x+3|-1</math>
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:05:56Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3) \approx 3.27</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10 \approx 4.32</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:01:42Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2}) = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T11:01:14Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above <math>| \frac{1}{2} -20 |</math> can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2} = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:59:07Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above | 1/2 -20 | can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2} = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

*| 2x + 1 |

*| x + 3| , if x < -3

*| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:57:10Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above | 1/2 -20 | can be simplified as follows:

<math>| \frac{1}{2} -20 | = | \frac{-39}{2} | = -(\frac{-39}{2} = \frac{39}{2}</math>

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:54:34Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above | 1/2 -20 | can be simplified as follows:

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples involving algebraic expressions are now presented

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

2x + 1 is always positive and according to the definition above | 2x + 1 | can be simplified as follows:

| 2x + 1 | = 22 + 1

If x < -3, then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows:

| x + 3 | = -(x + 3) = -x - 3

If x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows:

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:51:16Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows:

| -2 + 10 | = | 8 | = 8

<math>\frac{1}{2} - 20 = \frac{-39}{2}</math> is negative, according to the definition above | 1/2 -20 | can be simplified as follows:

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

<math>\sqrt(3) - 5</math> is approximately equal to -3.27 which is negative, according to the definition above <math>| \sqrt(3) - 5 |</math> can be simplified as follows:

<math>| \sqrt(3) - 5 | = - ( \sqrt(3) - 5) = 5 - \sqrt(3)</math>

<math>(\sqrt(14) - (3)(\pi) + 10</math>) is approximately equal to 4.32 which is positive, according to the definition above <math>| \sqrt(14) - (3)(\pi) + 10) |</math> can be simplified as follows:

<math>| \sqrt(14) - (3)(\pi) + 10 | = \sqrt(14) - (3)(\pi) + 10</math>

Examples with algebraic expressions are now presented.

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:44:17Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

*<math>|(-2) + 10|</math>

*<math>| \frac{1}{2} -20 |</math>

*<math>| \sqrt(3) - 5 |</math>

*<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:39:33Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

<math>|-2 + 10|</math>

<math>| \frac{1}{2} -20 |</math>

<math>| \sqrt(3) - 5 |</math>

<math>| \sqrt(14) - (3)(\pi) + 10 |</math>

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:35:02Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

'''Example 1:''' Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

'''Solution to Example 1'''

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

'''Example 2:''' Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

'''Solution to Example 2'''

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T10:33:58Z

JonathanRothwell: /* How to Use Absolute Value Functions */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.

Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:49:07Z

JonathanRothwell: /* What are they? */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>,

resulting in the graphs 'V' shaped graphs seen below.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:47:02Z

JonathanRothwell: /* What are they? */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

<math>|a|</math>

The definition of an absolute value function follows as:

<math>|a| =
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

Interestingly the absolute function is equivalent to the square root of <math>x^2</math>, such as in the following case:

<math>|x| = \sqrt{x^2} </math>

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[http://www.mathsisfun.com/sets/images/function-absolute.gif]

f(x) = |x| = x, if x > or equal to 0
f(x) = |x| = -x, if x < 0

Piecewise functions can typically occur in as many regions as possible (often times not more than 2 or 3) but in theory, we could have a piecewise graph that has 20 different regions and it would still be acceptable. So then what we do is graph the function in the region that is respective region that is given.

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:32:24Z

JonathanRothwell:

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

== Further Reading ==

# http://www.analyzemath.com/Absolute_Value_Function/Absolute_Value_Function.html
# http://www.analyzemath.com/Definition-Absolute-Value/Definition-Absolute-Value.html

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:26:17Z

JonathanRothwell: /* How to link them with piecewise functions? */

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to Link Them With Piecewise Functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:25:37Z

JonathanRothwell:

'''Basic Skills Project - Absolute Value Functions'''

==What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

==Graphing Absolute Value Functions==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

==How to Use Absolute Value Functions==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==How to link them with piecewise functions?==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:24:46Z

JonathanRothwell:

'''Basic Skills Project - Absolute Value Functions'''

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

==4. How to link them with piecewise functions? ==

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:22:12Z

JonathanRothwell:

'''Basic Skills Project - Absolute Value Functions'''

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

----

'''How to link them with piecewise functions?'''

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

----

== References ==

# http://www.vias.org/calculus/01_real_and_hyperreal_numbers_02_14.html
# http://www.tutorvista.com/math/absolute-value-rules

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:18:14Z

JonathanRothwell: /* 1. What are they? */

'''Basic Skills Project - Absolute Value Functions'''

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

'''Domain & Range'''

* Domain of absolute value functions is defined with all real numbers
* Range of absolute value functions is on the interval [0,<math>\infty</math>)

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

----

'''How to link them with piecewise functions?'''

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

[[File:f1q99g1.jpg]]

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

----

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:12:27Z

JonathanRothwell:

'''Basic Skills Project - Absolute Value Functions'''

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

----

'''How to link them with piecewise functions?'''

In real world situations it can be difficult to describe events with a single smooth function. This is why sometimes we need to use piecewise functions to describe these situations. Piecewise functions are functions that are defined to be smooth functions for specific intervals of the independent variable (most commonly the x-variable).

The link below shows a video which goes over the concept of a piecewise function and provides two problems.

http://www.brightstorm.com/math/algebra-2/additional-topics/piecewise-functions

The Absolute Value Function is a famous Piecewise Function.

It has two pieces:

below zero: -x
from 0 onwards: x

f1q99g1.jpg

f(x) = |x| = x, if x > or equal to 0
-x, if x < 0

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:08:11Z

JonathanRothwell: /* Basic Skills - Absolute Value Functions */

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:07:47Z

JonathanRothwell: /* Basic Skills */

=='''Basic Skills - Absolute Value Functions'''==

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T08:07:03Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

==1. What are they?==

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----

== 2. Graphing Absolute Value Functions ==

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

== 3. How to Use Absolute Value Functions ==

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:56:06Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----
'''2. Graphing Absolute Value Functions'''

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''3. How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:55:57Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----
'''2.Graphing Absolute Value Functions'''

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''3. How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:45:15Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =
<math>
\begin{cases}
a, & \mbox{if }a\mbox{ is greater than or equal to zero} \\
-a, & \mbox{if }a\mbox{ is less than zero}
\end{cases}</math>

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:35:41Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of the function:

|a|

The definition of an absolute value function follows as:

|a| =

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:35:20Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of function:

|a|

The definition of an absolute value function follows as:

|a| =

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:34:30Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

An absolute value is represent with a vertical bar on either side of function:

|a|

The definition of an absolute value function follows as:

|a| =

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:22:50Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

|a| ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T07:16:30Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

----

'''How to Use Absolute Value Functions'''

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T06:58:59Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

How to Use Absolute Value Functions.

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T06:58:49Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

\begin{cases}
3x + 5y + z \\
7x - 2y + 4z \\
-6x + 3y + 2z
\end{cases}

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

How to Use Absolute Value Functions.

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T06:56:38Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

How to Use Absolute Value Functions.

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T06:41:43Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

How to Use Absolute Value Functions.

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03/Basic Skills Project

2010-12-03T06:36:39Z

JonathanRothwell: /* Basic Skills */

==Basic Skills==

'''Absolute Value Functions'''

'''1. What are they?'''

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition of an absolute value function follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

----
Graphing Absolute Value Functions

Taking the absolute value of a negative number makes it positive. For this reason, graphs of absolute value functions tend not to look quite like the graphs of linear functions that we have already studied. Because of how absolute values behave, it is important to include negative inputs in our T-chart when graphing absolute-value functions. If we do not pick x-values that will put negatives inside the absolute value, we usually mislead ourselves as to what the graph is.While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. In all cases, you should take care that you pick a good range of x-values; three x-values right next to each other will almost certainly not give you anywhere near enough information to draw a valid picture.
http://www.youtube.com/watch?v=h2ofpYSBR1c

How to modify Absolute values
The parent graph of an absolute value graph is y=|x|, which looks like :
http://i38.photobucket.com/albums/e148/supperz/graph1.jpg

But how about a Function like -1/2|x+3|-1? *When you have a function in the form y = |x + h| the graph will move h units to the left.
*When you have a function in the form y = |x - h| the graph will move h units to the right.
*When you have a function in the form y = |x| + k the graph will move up k units.
*When you have a function in the form y = |x| - k the graph will move down k units.
*If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.

In this case, when graphing you begin with what inside the absolute value sign.

1)You start off with what is inside the absolute value sign.

Shift the graph three units to the left y=|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph2.jpg

2)The next step is to take care of the -1/2 coefficient y= -1/2|x+3|

The negative sign is a reflection on the x axis, and the half is a vertical expansion of 2.
*-|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph3.jpg

*-1/2|x+3|
http://i38.photobucket.com/albums/e148/supperz/graph4.jpg

3)The last step is to shift the graph one unit down -1/2|x+3|-1
http://i38.photobucket.com/albums/e148/supperz/graph5.jpg

How to Use Absolute Value Functions.

Simplifying Absolute Value Functions.

If x > = 0 then | x | = x and if x < 0 then | x | = -x.

Follow the examples below and try to understand each step.
Example 1: Simplify the expressions with absolute value

| -2 + 10 |

| 1/2 -20 |

| sqrt(3) - 5 |

| sqrt(14) - 3*pi + 10 |

Solution to Example1

- 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows

| -2 + 10 | = | 8 | = 8

1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows

| 1/2 -20 | = | -39/2 | = -(-39/2) = 39/2

sqrt(3) - 5 is approximately equal to -3.27 which is negative, according to the definition above | sqrt(3) - 5 | can be simplified as follows

| sqrt(3) - 5 | = - ( sqrt3 - 5) = 5 - sqrt(3)

sqrt(14) - 3*pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | sqrt(14) - 3*pi + 10 | can be

simplified as follows

| sqrt(14) - 3*pi + 10 | = sqrt(14) - 3*pi + 10

Examples with algebraic expressions are now presented.

Example 2: Simplify the algebraic expressions with absolute value

| 2x + 1 |

| x + 3| , if x < -3

| -x + 2 | , if x > 2

Solution to Example 2

x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows

| x2 + 1 | = x2 + 1

if x < -3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows

| x + 3 | = -(x + 3) = -x - 3

if x > 2 then x - 2 > 0 and -x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows

| -x + 2 | = - ( - x + 2 ) = x – 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-11-29T11:21:39Z

JonathanRothwell: /* Basic Skills */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 5 =
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

JON - Don't Understand
*Domain of radicals (especially writing interval notation, at least for WeBWork)
*polynomial long division
*Intersection of functions
*Limited understanding if trig

* polynomials
*
What I understand- Supreet:-
*Basic Funtions
*Property of functions
*Equations
*Inequalties
*Composite Functions
* Trigonometry and the Pythagorean theorem
* Areas and volumes
*polynomial long division

What I don't Understand:-
*Reading Graph of Functions
*Distances of Lines
*Mathematical writing
*Intersection of functions
*Graph of Functions
*Construction of graphs

What I Understand-Ghita Youssefi

Properties of functions
Equations
Inequalities
Composition of functions
Distances and lines
Trigonometry and the Pythagorean theorem
Areas and volumes
Mathematical writing

What I Don't Understand- Ghita Youssefi:

Properties of functions
Composition of functions
Polynomial long division
Graphs of functions
Intersections of functions
Reading graphs of functions
Operations on graphs of functions
Construction of graphs

''SOME'' of us understand:
Properties of functions
Composition of functions
Polynomial long division

''NONE'' of us understand:
polynomial long division
Intersection of functions
Operations on graphs of functions

Daisy Gobina
I understand most of the topics on the basic skills test but i think i have forgotten how to go about some things e.g i dont really remember Trigonometry and i think i really do bad at it. I also dont know volumes and areas of figures execpt those i think am ok with the others realatively. That is i understand what is happening

=Basic Skills=

Absolute Value Functions

1. What are they?

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition follows as:

abs(a) ={a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-11-29T11:21:23Z

JonathanRothwell: /* Basic Skills */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 5 =
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

JON - Don't Understand
*Domain of radicals (especially writing interval notation, at least for WeBWork)
*polynomial long division
*Intersection of functions
*Limited understanding if trig

* polynomials
*
What I understand- Supreet:-
*Basic Funtions
*Property of functions
*Equations
*Inequalties
*Composite Functions
* Trigonometry and the Pythagorean theorem
* Areas and volumes
*polynomial long division

What I don't Understand:-
*Reading Graph of Functions
*Distances of Lines
*Mathematical writing
*Intersection of functions
*Graph of Functions
*Construction of graphs

What I Understand-Ghita Youssefi

Properties of functions
Equations
Inequalities
Composition of functions
Distances and lines
Trigonometry and the Pythagorean theorem
Areas and volumes
Mathematical writing

What I Don't Understand- Ghita Youssefi:

Properties of functions
Composition of functions
Polynomial long division
Graphs of functions
Intersections of functions
Reading graphs of functions
Operations on graphs of functions
Construction of graphs

''SOME'' of us understand:
Properties of functions
Composition of functions
Polynomial long division

''NONE'' of us understand:
polynomial long division
Intersection of functions
Operations on graphs of functions

Daisy Gobina
I understand most of the topics on the basic skills test but i think i have forgotten how to go about some things e.g i dont really remember Trigonometry and i think i really do bad at it. I also dont know volumes and areas of figures execpt those i think am ok with the others realatively. That is i understand what is happening

=Basic Skills=

Absolute Value Functions

1. What are they?

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition follows as:

abs(a) ={a if a > or equal to 0 -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-11-29T11:14:06Z

JonathanRothwell: /* Basic Skills */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 5 =
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

JON - Don't Understand
*Domain of radicals (especially writing interval notation, at least for WeBWork)
*polynomial long division
*Intersection of functions
*Limited understanding if trig

* polynomials
*
What I understand- Supreet:-
*Basic Funtions
*Property of functions
*Equations
*Inequalties
*Composite Functions
* Trigonometry and the Pythagorean theorem
* Areas and volumes
*polynomial long division

What I don't Understand:-
*Reading Graph of Functions
*Distances of Lines
*Mathematical writing
*Intersection of functions
*Graph of Functions
*Construction of graphs

What I Understand-Ghita Youssefi

Properties of functions
Equations
Inequalities
Composition of functions
Distances and lines
Trigonometry and the Pythagorean theorem
Areas and volumes
Mathematical writing

What I Don't Understand- Ghita Youssefi:

Properties of functions
Composition of functions
Polynomial long division
Graphs of functions
Intersections of functions
Reading graphs of functions
Operations on graphs of functions
Construction of graphs

''SOME'' of us understand:
Properties of functions
Composition of functions
Polynomial long division

''NONE'' of us understand:
polynomial long division
Intersection of functions
Operations on graphs of functions

Daisy Gobina
I understand most of the topics on the basic skills test but i think i have forgotten how to go about some things e.g i dont really remember Trigonometry and i think i really do bad at it. I also dont know volumes and areas of figures execpt those i think am ok with the others realatively. That is i understand what is happening

=Basic Skills=

Absolute Value Functions

1. What are they?

Absolute values are the numerical value of a real number whereby positive and negative signs are disregarded.

The definition follows as:

abs(a) = a if a > or equal to 0; -a if a < 0

Therefore the solution of an absolute value is always greater than or equal to zero.

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-11-29T11:06:23Z

JonathanRothwell:

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-10-29T02:01:10Z

JonathanRothwell: /* Homework 5 part 3 */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:YuriTomura|Yuri Tomura]]
| member 5 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

JON - Don't Understand
*Domain of radicals (especially writing interval notation, at least for WeBWork)
*polynomial long division
*Intersection of functions
*Limited understanding if trig

''ALL'' of us understand:
* polynomials
*

''SOME'' of us understand:
*

''NONE'' of us understand:
*

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-10-29T02:00:24Z

JonathanRothwell: /* Homework 5 part 3 */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:YuriTomura|Yuri Tomura]]
| member 5 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

JON - Don't Understand
*Domain of radicals
*polynomial long division
*Intersection of functions
*Limited understanding if trig

''ALL'' of us understand:
* polynomials
*

''SOME'' of us understand:
*

''NONE'' of us understand:
*

Course:MATH110/Archive/2010-2011/003/Groups/Group 03

2010-10-29T01:59:16Z

JonathanRothwell: /* Homework 5 part 3 */

{{Infobox MATH110 Groups
| group number = 3
| member 1 = Daisy Gobina
| member 2 = Jonathan Rothwell
| member 3 = Supreet Saran
| member 4 = [[User:YuriTomura|Yuri Tomura]]
| member 5 = [[User:GhitaYoussefi|Ghita Youssefi]]
| member 6 =
}}

=Homework 3 part 3=

===Questions 1-10 ''Yuri Tomura''===

1. There is actually no time difference. It is just the fact that the time is written differently. It says the time it took for the bus to travel from the terminal to the airport was an hour and 20 mins which is the same as 60min + 20min = 80 mins, hence the same time duration it took for the bus to travel back to the terminal from the airport.

2. Although the question mentions a driver's license, we should not assume that this lady was driving a car. She could have been walking in which case she would not be breaking any laws - it is fine for her to not have her license, the stop sign is only for driver's and a one-way street does not apply to pedestrians. However, if she was driving a car, we could also come up with the fact that the policeman didn't see her when she broke the law but only saw her when she was actually obeying the rules.

3. The question mentions how the boxes are labeled APPLES, ORANGES and APPLES AND ORANGES. Therefore, we can lay this out as:

Box 1 --> APPLES

Box 2 --> ORANGES

Box 3 --> APPLES AND ORANGES

However, all the boxes have the incorrect labels.

Our task is to draw one fruit from only one box and determine the correct labels.

If all the labels are incorrect, we will end up with:

Box 1 --> ORANGES or APPLES AND ORANGES

Box 2 --> APPLES or APPLES AND ORANGES

Box 3 --> either APPLES or ORANGES

So, if we select a fruit from box 3 and it is an apple, then box 2 is a box with apples and oranges as we can eliminate the option of only apples. Therefore, Box 1 would contain just oranges as there no longer can be the option of apples and oranges. Consequently, it is possible to select one fruit from only one box and determine the correct labels.

4. Looking at this question, we can see that the gender of the "blind fiddler" is not specified. The question mentions that "I am the brother of the blind fiddler" so we know that "I" is a male. Futhermore, we can figure out that the blind fiddler is a female as he has no brothers which concludes us to the fact that the blind fiddler is his sister.

5. Both coins are quarters, meaning they have the same dimensions. Therefore, when it revolves once around it, it will only be halfway around the quarter. So, the coin would have to revolve twice to go back to its original position.

6. There are 3 kinds of apples: Apple a, b and c.

First pick --> Apple a

Second pick --> Apple b

Third pick --> Apple c

Fourth pick --> Apple a

Now we have two of the same kind of apple a. Therefore, we need to draw an apple at least 4 times t in order to get at least two of the same kind.

7. i) In order to be sure to pick a pair of the same colour socks, we must take 3 socks from the drawer since if only two socks were picked, there is a possibility that they are different colours. But, if we pick 3, then at least 2 of them will bound to be the same colour.
ii) In order to pick a pair of different colour socks, we must also take 2 socks from the drawer since if only two socks were picked, there is a possibility that they are both the same colours but if we pick 3, then at least 2 of them will be different colours.

8. Reuben's birthday is on December 31st and he said "Two days ago I was 20 years old. Later next year I will be 23 years old" on January 1st. This means he ''was'' 20 years old on December 30th but turned 21 on December 31st last year. This year he will be turning 22 and so later next year, he will be 23 years old.

9. If the tide rises five feet, the boat would rise five feet as well. Therefore, the rope ladder hanging over the side of the boat would also rise five feet, leaving 10 rungs to be showing still.

10. i) No, it does not follow that one-fourth of all people are women chocolate eaters because if half of all people are chocolate eaters and half of all people are women, it does not necessary mean that one-fourth of all people are women chocolate eaters as there are many other possibilities of which half of the people are chocolate eaters.

=Homework 4 part 3=
Question 1- Ghita Youssefi

A table was made in order to break up the problem into three categories: Owners, Pets, and Pets Names.

'''Owners:'''

Tosh --> (male)
Bianca --> (female)
Jaela ---> (female)
Jun --> (male)
Susan --> (female)

'''Pets Names:'''

Cat --> Tosh
Frog --> Bianca
Turtle --> Susan
Parrot --> Jaela
Snake --> Jun
In order to find who the turtle belonged to, I eliminated all the other owners and Susan was the only left.

'''Pets:'''

Tosh --> Parrot
Bianca --> Snake
Jaela --> Turtle
Jun --> Cat (provided)
Susan --> Frog (provided)

We are told that the name by which they call the turtle is the name of the woman is Tosh, therefore, Tosh's owner is female and Bianca's pet is Susan. In consequence, Tosh has to be the Parrot and Jaela has to be the turtle. Finally, the only pet left is the snake and it is paired up with the last pet name which is Bianca.

The last step is finding out which pet belong's to Susan's mother. We know that Susan's pet is the turtle called Jaela. Also, the cat, Jun, belongs to Susan's (the frog) Mother (Bianca).

Problem 2 (Supreet)
In this problem, we have five different basketball players. We are trying to find the center player who is left handed, and over 2m tall. Using Pólya’s method, our first step to understand the problem. The conditions we know are that two of the players are left handed and three right are handed. Bohao and Dylan use the same hand, white Tim and Chan use different hands. The condition about height we know is that two are over 2m tall and three are under 2m tall. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges
The second step in Pólya’s method is to plan a strategey. My strategey in this case is to use a chart, and use the process of elimation. The third step is to execute it.
Firstly, we know that that theres 2/5 left handed, which mean Bohao and Dylan must be right-handed, as Tim and Chan are 2 '''different''' hands. If Bohao & Dylan were left handed, then there would be 3 left-handed, as either Tim and Chan would be left handed as well, which is a contradiction to the condition. Therefore, they are eliminated from being in centre position.
Secondly, we also kown there are 2/5 over 2m tall. This is similar to the our reasoning above, as we know stewart and chan are the same height, when Dylan and Tim are different from each other, which means that stewart and chan must be under 2m. Dylan and Tim could be either or.
The only one who hasn't been eliminated in both conditions it Tim.
Our fourth step is to check and interpret our steps.Bohao, Dylan, Stewart, and Chan are elimanted as their conditions don't comply with the left-hand, and above 2m conditions of the centre player. Consequently, this leaves Tim. Tim is who we were looking for, he is left handed and over 2m. Therefore, Tim is centre player.
http://i38.photobucket.com/albums/e148/supperz/math110.png

'''Problem 4''' ''Yuri Tomura''

Firstly, I made a table that looked like this and filled out the known information given in the question:
[[File:table 1.jpg]]

Then, I used Petra's column to fill up the blanks by trial and error. I started of with Petra playing against Fernanda on the third day:
[[File:table 2.jpg]]

Now we have Petra yet left to play with Janet and Li on the second and fifth day. However, we can see that Janet is already playing Carla on the second day which means Petra cannot play Janet on the second day, hence, Petra played Li on the second day and Janet on the fifth:
[[File:table 3.jpg]]

Next we can see that Janet only has either the first day or fourth day left to play against Sandra or Fernanda. It is obvious that Sandra is already playing on the fourth day against Petra so we can conclude that Janet will be playing Sandra on the first day and Fernanda on the fourth:
[[File:table 4.jpg]]

Now looking at Fernanda's column, we can see that she still needs to play on the first, second and fifth day against Carla, Sandra and Li. However, it is clear that Fernanda '''cannot''' play Carla or Li on the second day as they are playing on that day already. This leaves us with Fernanda playing against Sandra on the second day:
[[File:table 5.jpg]]

Finally, since Carla is already playing Petra on the first day from the table above, it results as Fernanda playing against Li on the first day, thus leaving Fernanda to play Carla on the fifth day. Therefore, the answer is Carla:
[[File:table 6.jpg]]

'''Problem 5''' (Jon)

Conditions:

*(a) No one of the noisemakers were quiet for 3 consecutive days
*(b) No pair of the noisemakers made noise on more than 1 day

Possible combinations of noisemakers include:

*Salesman & Dog
*Salesman & Construction
*Construction & Dog

The provided information in the problem is:

*Monday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday -
*Friday -
*Saturday - construction
*Sunday - dog

Therefore, dog barking must wake Homer on Saturady since no noisemaker can be quiet for 3 consecutive days and the previous day was Wednesday. This completes the combinations leaving on single noisemakers to disturb Homer. Between Wednesday and Sunday the salesman must occur on Friday since he can't occur on Saturady due to the contruction and the dog, therefore he must occur Friday in order to obey condition (a). Construction must once again occur on Thursday since the previous day for construction was Monday.

Determined Days:

*Saturday - salesman
*Sunday - dog
*Monday - salesman & construction
*Tuesday -
*Wednesday - salesman & dog
*Thursday - '''CONSTRUCTION'''
*Friday - '''SALESMAN'''
*Saturday - construction & '''DOG'''
*Sunday - dog

The only remaining day whereby no violoations of the conditions occur and where Homer is able to sleep in is Tuesday.

Problem 4(Daisy)

Fristly I said Clara played Petra on the first day.Jannet played Sandra and Li played Fernanda. This was because each of them was to play each ther once during the competion. On the second day Jannet will play clara as we are told in the ouestion and so Petra will play Li since we are told on the fourth day Sandra will play Petra that means they can't play now since they are too meet each other just once in the competition. On the second day Sandra will paly Li. On the third day We are told that Li played Jannet, So since Clara has already played Petra and Jannet so she is eligible to play only Sandra ,Li and Fernanda and so since Sandra is still to play Petra , Clara can only play Sandra and Petra will play Fernanda.On the fourth day Petra plays Sandra and Since Li has already played Jannet she cannot play her again so Jannet will play Fernanda and Li will play Clara. On the Fourth day Since Petra has played Clara,Li,Fernanda Sandra the only person she can now play on the fifth day is Jannet. Since Sandra has played Jannet,Fernanda,Clara and Petra the only person she can now play is Li. Eventually Clara and Fernanda are left so they are going to play against each other . So on the fifth day Clara is going to play Fernanda

=Homework 5 part 3=

'''What I understand - Yuri'''
* Basic Functions
* Properties of functions
* Equations
* Graphs of functions
* Polynomial long division
* Intersections of functions
* Distances and lines
* Construction of graphs
* Trigonometry and the Pythagorean theorem
* Areas and volumes
* Mathematical writing

What I don't - Yuri
* Inequalities
* Composition of funcitons
* Operations on graphs of functions

'''JON - Understand'''
*Basic functions
*Properties of functions
*Equations
*Inequalities
*Composite functions
*Graphs of functions
*Readings graphs of functions
*Distances & lines
*Operations on graphs of functions
*Construction of graphs
*Areas & volumes
*Math writing
*Pythagorean theorem

Don't Understand
*Domain of radicals
*polynomial long division
*Intersection of functions
*Limited understanding if trig

''ALL'' of us understand:
* polynomials
*

''SOME'' of us understand:
*

''NONE'' of us understand:
*