Answer
(a) $ x= 3$ is a vertical asymptote
(b) $ x= 1$ is a vertical asymptote
(c) $ x=-4$ is a vertical asymptote
Work Step by Step
(a) Since $ y=\frac{x^{2}+4}{x-3}$ is undefined at $ x=3$ :
$ \lim _{x \rightarrow 3^{-}} \frac{x^{2}+4}{x-3}=-\infty $ and $\lim _{x \rightarrow 3^{+}} \frac{x^{2}+4}{x-3}=+\infty$
Thus, $ x=3$ is a vertical asymptote.
(b) $ y=\frac{x^{2}-x-2}{x^{2}-2 x+1}$ is undefined at $ x=1 $:
$ \lim _{x \rightarrow 1} \frac{x^{2}-x-2}{x^{2}-2 x+1}=-\infty $ and $\lim _{x \rightarrow 1^{+}} \frac{x^{2}-x-2}{x^{2}-2 x+1}=-\infty$
Thus, $ x=1$ is a vertical asymptote.
(c) Since $ y=\frac{x^{2}+x-6}{x^{2}+x-8}$ is undefined at $ x=2$ and $-4$:
$\begin{align*}\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x^{2}+2 x-8}&=\lim _{x \rightarrow 2} \frac{x+3}{x+4}=\frac{5}{6}\\
\lim _{x \rightarrow-4} \frac{x^{2}+x-6}{x^{2}+2 x-8}&=\lim _{x \rightarrow-4^{-}} \frac{x+3}{x+4}=\infty\end{align*}$
$\lim _{x \rightarrow-4^{+}} \frac{x^{2}+x-6}{x^{2}+2 x-8}=\lim _{x \rightarrow-4^{+}} \frac{x+3}{x+4}=-\infty .$ Thus, $ x=-4$ is a vertical asymptote.
