Question

Find the vertical asymptotes, if any, and the horizontal or oblique asymptote, if any, of$y=\frac{x^2-9}{x^2-4}$

Short answer

The equation $y=\frac{x^2-9}{x^2-4}$has vertical asymptotes at $x=2$and $x=-2.$

The equation $y=\frac{x^2-9}{x^2-4}$has horizontal or oblique asymptotes at$y=1$.

Step-by-step solution

Step 1. Given Information

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

Step 2. Aim of the problem

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

Step 3. Concept

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 $f\left(x\right)=ax+b$, $a\ne 0$, 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.

Step 4. Finding the vertical asymptote

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 $x.$

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

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

Step 5. Finding Horizontal asymptotes

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$y=1$.