Chapter 5.3, Problem 16E

Solution

To graph : The given function.

Given information :

Function $y=\frac{8}{x^2-x-6}$

Graph :

  

Interpretation :

Since $\frac{8}{\left(x-3\right)\left(x+2\right)}\to \infty$ as $x\to -2$ from the left and $\frac{8}{\left(x-3\right)\left(x+2\right)}\to -\infty$ as $x\to -2$ from the right then  $x=-2$  is a vertical asymptote.

Since $\frac{8}{\left(x-3\right)\left(x+2\right)}\to -\infty$ as $x\to 3$ from the left and $\frac{8}{\left(x-3\right)\left(x+2\right)}\to \infty$ as $x\to 3$ from the right then  $x=3$  is a vertical asymptote.

List all of the vertical asymptotes:

  $x=-2,3$

Since  $n<m$ , the x-axis,  $y=0$ , 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:  $x=-2,3$

Horizontal Asymptotes:  $y=0$

No Oblique Asymptotes