Answer
The equations of the vertical asymptotes of the functions are:
a) $x=3$
b) $x=1$
c) $x=-4$
Work Step by Step
a) $$y=\frac{x^2+4}{x-3}$$
- For $x\to3^+$, $(x-3)\to0^+$, meaning $(x-3)\gt0$, while $(x^2+4)\to13^+\gt0$. Therefore, $\lim_{x\to3^+}\frac{x^2+4}{x-3}=\infty$
- For $x\to3^-$, $(x-3)\to0^-$, meaning $(x-3)\lt0$, while $(x^2+4)\to13^-\gt0$. Therefore, $\lim_{x\to3^-}\frac{x^2+4}{x-3}=-\infty$
According to definition, $x=3$ is the vertical asymptote of the function.
b) $$f(x)=\frac{x^2-x-2}{x^2-2x+1}=\frac{(x-2)(x+1)}{(x-1)^2}$$
- For $x\to1^+$, $(x-1)^2\to0^+\gt0$, while $x^2-x-2\to-2\lt0$. Therefore, $\lim_{x\to1^+}\frac{x^2-x-2}{(x-1)^2}=-\infty$
- For $x\to1^-$, $(x-1)^2\to0^+\gt0$, while $x^2-x-2\to-2\lt0$. Therefore, $\lim_{x\to1^-}\frac{x^2-x-2}{(x-1)^2}=-\infty$
According to definition, $x=1$ is the vertical asymptote of the function.
c) $$y=\frac{x^2+x-6}{x^2+2x-8}=\frac{(x-2)(x+3)}{(x-2)(x+4)}$$
We have for $c\in R$, $$\lim_{x\to c}y=\lim_{x\to c}\frac{(x-2)(x+3)}{(x-2)(x+4)}=\lim_{x\to c}\frac{x+3}{x+4}$$
- For $x\to-4^+$, $(x+4)\to0^+\gt0$, while $(x+3)\to-1^+\lt0$. Therefore, $\lim_{x\to-4^+}y=\lim_{x\to-4^+}\frac{x+3}{x+4}=-\infty$
- For $x\to-4^-$, $(x+4)\to0^-\lt0$, while $(x+3)\to-1^-\lt0$. Therefore, $\lim_{x\to-4^-}y=\lim_{x\to-4^-}\frac{x+3}{x+4}=\infty$
According to definition, $x=-4$ is the vertical asymptote of the function.
