91Ó°ÊÓ

The given equation is $y=\frac{x^2-9}{x^2-4}$.

We need to find the vertical asymptotes if any and horizontal asymptotes if any of the given function.

Vertical Asymptote:

A rational function $R\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$,in lowest terms, will have a vertical asymptoteat $x=r$if $r$is a real zero of the denominator $q\left(x\right)$.

Proper Function:

A proper function is a rational function whose degree of the numerator is less than the degree of the denominator.

Horizontal Asymptote:

When the function is not a proper function, write the rational function as the sum of a polynomial function $f\left(x\right)$plus a proper rational function $\frac{r\left(x\right)}{q\left(x\right)}$using long division.

That is, we write $R\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=f\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)}$.

Then, we have the following cases:

1. If $f\left(x\right)=b$, a constant, then the line $y=b$ is a horizontal asymptoteof the graph of $R$.

2. If $f\left(x\right)=ax+b$, $a≠0$, then the line $y=ax+b$ is an horizontal asymptote of

the graph of $R$.

3. In all other cases, the graph of $R$approaches the graph of $f$, and there are no

horizontal or oblique asymptotes.

To find the vertical asymptote, we must find the zero of the denominator.

Note that zero of a function $y=f\left(x\right)$, is that value of $x$, for which the output, $y=0$.

The denominator of the given rational function is, $y=x^2-4$.

To find the zero of $y=x^2-4,$substitute $y=0$and then solve for $x.$

$$\begin{gathered} y=x^2-4 \\ 0=x^2-4 \\ {x}^2=4 \\ x=\sqrt{4} \\ x=Â\pm 2 \end{gathered}$$

Therefore, the lines $x=2$and $x=-2$are the vertical asymptotes of the given function.

The given rational function is not a proper function as the degree of the numerator is equal to the degree of the denominator.

Hence, we express the given function as the sum of a function $f\left(x\right)$and a proper function $\frac{r\left(x\right)}{q\left(x\right)}$by proper division.

Divide $x^2-9$by $x^2-4,$by long division.

Doing so, we get,

$\frac{x^2-9}{x^2-4}=1-\frac{5}{x^2-4}$

Here, $f\left(x\right)=1$

Hence, the given equation has a horizontal asymptote at$y=1$.