Graph f(x)=1/(4-x^2)

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

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

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

1. If ${\displaystyle n<m}$, then the x-axis, ${\displaystyle y=0}$, is the horizontal asymptote.

2. If ${\displaystyle n=m}$, then the horizontal asymptote is the line ${\displaystyle y=\frac{a}{b}}$.

3. If ${\displaystyle n>m}$, then there is no horizontal asymptote (there is an oblique asymptote).

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.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes: ${\displaystyle x=-2,2}$

Horizontal Asymptotes: ${\displaystyle y=0}$

No Oblique Asymptotes