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

MATH 180 December 2017

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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.

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 ${\textstyle \lim \limits _{x\to c^{+}}f(x)=-\infty }$.

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.

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

Checking a solution serves two purposes: helping you if, after having used all the hints, 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. 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 ${\textstyle {\color {blue}x=-3}}$.