Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 03 (a)

MATH 180 December 2017

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[hide] Question 03 (a)
(a) Let $f(x)={\frac {x^{2}+x-2}{x^{2}+2x-3}}$. Find and justify the vertical asymptotes of ${\textstyle f(x)}$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
The vertical asymptotes are the vertical lines ${\textstyle x=c}$, where either ${\textstyle \lim \limits _{x\to c^{-}}f(x)=+\infty }$, ${\textstyle \lim \limits _{x\to c^{+}}f(x)=+\infty }$, ${\textstyle \lim \limits _{x\to c^{-}}f(x)=-\infty }$ or ${\textstyle \lim \limits _{x\to c^{+}}f(x)=-\infty }$.

[show] Hint 2
The candidates of ${\textstyle c}$ are the roots of the quadratic equation ${\textstyle x^{2}+2x-3=0}$. Note that the left hand side can be factorized into two linear factors.

[show] Hint 3
The numerator can also be factorized. Pay attention to the factor that can be cancelled.

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[show] Solution

Let us first simplify the function: $$f(x)={\dfrac {x^{2}+x-2}{x^{2}+2x-3}}={\dfrac {(x+2)(x-1)}{(x+3)(x-1)}}.$$ Note that ${\textstyle f(x)}$ is defined for all real numbers except at ${\textstyle -3}$ (where ${\textstyle x+3=0}$) and ${\textstyle 1}$ (where ${\textstyle x-1=0}$). The candidates for vertical asymptotes are ${\textstyle x=-3}$ and ${\textstyle x=1}$. However we need to compute the limits of ${\textstyle f(x)}$ at these values to check whether each is a vertical asymptote. For ${\textstyle x\neq -3,1}$, we can write $$f(x)={\dfrac {x+2}{x+3}}.$$ From this we see that $$\lim _{x\to -3^{+}}f(x)=\lim _{x\to -3^{+}}{\dfrac {x+2}{x+3}}=-\infty ,\quad \quad \lim _{x\to -3^{-}}f(x)=\lim _{x\to -3^{-}}{\dfrac {x+2}{x+3}}=\infty$$ and $$\lim _{x\to 1}f(x)=\lim _{x\to 1}{\dfrac {x+2}{x+3}}={\dfrac {\lim \limits _{x\to 1}(x+2)}{\lim \limits _{x\to 1}(x+3)}}={\dfrac {1+2}{1+3}}={\dfrac {3}{4}}.$$ Therefore, ${\textstyle x=-3}$ is a vertical asymptote, while ${\textstyle x=1}$ is not one. Answer: The (only) vertical asymptote is ${\textstyle {\color {blue}x=-3}}$.