Chapter 9, Problem 33SGR

To determine

The equations of any vertical asymptotes and the values of x for ny holes in the graph of rational function.

Answer to Problem 33SGR

The vertical asymptotes is $x=8$ and the hole is at $x=-3$ .

Explanation of Solution

Given information:

The given rational function is $f\left(x\right)=\frac{x^2-9}{x^2-5x-24}$ .

Calculation:

If $f\left(x\right)=\frac{a\left(x\right)}{b\left(x\right)}$ , $a\left(x\right)$ and $b\left(x\right)$ are polynomial function with no common factor other than 1, and $b\left(x\right)\ne 0$ , then $f\left(x\right)$ has vertical asymptotes whenever $b\left(x\right)=0$ .

And if $x-c$ is the factor of both $a\left(x\right)$ and $b\left(x\right)$ , then there is a point of discontinuity at $x=c$ .

Therefore, for the function $f\left(x\right)=\frac{x^2-9}{x^2-5x-24}$ .

  $\begin{array}{l}f\left(x\right)=\frac{x^2-9}{x^2-5x-24}\\ f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-8\right)\left(x+3\right)}\end{array}$

So the function is undefined for $x=8$ and $x=-3$ .

Since,

  $\begin{array}{l}f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-8\right)\left(x+3\right)}\\ f\left(x\right)=\frac{\left(x-2\right)}{\left(x-8\right)}\end{array}$

So, $x=8$ is a vertical asymptotes and $x=-3$ represents a hole in the graph.

Hence,

The vertical asymptotes is $x=8$ and the hole is at $x=-3$ .