Graph f(x)=(x^2+x-6)/(x^2+x-20)

Find where the expression ${\displaystyle \frac{(x-2)(x+3)}{(x-4)(x+5)}}$ is undefined.

Since ${\displaystyle \frac{(x-2)(x+3)}{(x-4)(x+5)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -5}$ from the left and ${\displaystyle \frac{(x-2)(x+3)}{(x-4)(x+5)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -5}$ from the right, then ${\displaystyle x=-5}$ is a vertical asymptote.

Since ${\displaystyle \frac{(x-2)(x+3)}{(x-4)(x+5)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 4}$ from the left and ${\displaystyle \frac{(x-2)(x+3)}{(x-4)(x+5)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 4}$ from the right, then ${\displaystyle x=4}$ is a vertical asymptote.

List all of the vertical asymptotes:

Consider the rational function ${\displaystyle R\left(x\right)=\frac{a{x}^n}{b{x}^m}}$ where ${\displaystyle n}$ is the degree of the numerator and ${\displaystyle m}$ is the degree of the denominator.

Find ${\displaystyle n}$ and ${\displaystyle m}$.

Since ${\displaystyle n=m}$, the horizontal asymptote is the line ${\displaystyle y=\frac{a}{b}}$ where ${\displaystyle a=1}$ and ${\displaystyle b=1}$.

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

This is the set of all asymptotes.