Science:Math Exam Resources/Courses/MATH110/December 2011/Question 07 (e)

MATH110 December 2011

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Question 07 (e)
Let $f(x)=x^{2}-6x$ and let P be the point (4, -12). (e) Sketch the curve $y=f(x)$, the point P, and the lines whose slopes you found in part (d).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
To draw the function $y=f(x)$: first try finding the roots by setting the function equal to zero and solving for x. Where will the minimum of the function be? To draw the lines: to draw lines you need a slope and a point. What are the slopes of these lines? Which point do they go through?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. $f(x)=x^{2}-6x$ As suggested in the hint, first find the roots of the function. $x^{2}-6x=0$ $x(x-6)=0$ $f(x)$ has roots at x = 0 and x = 6. Using this fact, or by calculating where the derivative of $f$ is zero, we can deduce that the minimum of $f(x)$ is at x = 3. To find the y-value at this point we simply plug in x = 3 to get $y=(3)^{2}-6(3)=-9$. Plotting the roots and the minimum on the graph should allow you to sketch the curve. $P=(4,-12)$ Tangent lines that pass through P We know there are two lines. Both pass through the point $(4,-12)$ and their two slopes are -2 and 6 (from part c). The equation of the two lines are thus $y+12=-2(x-4)$ $y+12=6(x-4)$ Or in y = mx + b form $y+12=-2x-4$ $y=6x-36$ Alternatively, you know that the line with slope -2 passes through P and the point (2, f(2)) = (2, -8), while the line with slope 6 passes through P and the point (6, f(6)) = (6, 0). Drawing these points and then connecting them with lines will also produce the correct picture.

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