Graph F(x)=1/((x+1)(x-2)(x-3))

Find where the expression ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$ is undefined.

Since ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -1}$ from the left and ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -1}$ from the right, then ${\displaystyle x=-1}$ is a vertical asymptote.

Since ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 2}$ from the left and ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 2}$ from the right, then ${\displaystyle x=2}$ is a vertical asymptote.

Since ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 3}$ from the left and ${\displaystyle \frac{1}{(x+1)(x-2)(x-3)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 3}$ from the right, then ${\displaystyle x=3}$ 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 x-axis, ${\displaystyle y=0}$, is the horizontal asymptote.

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.