Chapter 8, Problem 30CT

Solution

To identify:The function whose graph has a hole a vertical asymptote and a horizontal asymptote.

Explanation:

Consider the below rational function,

  $y=\frac{\left(x+6\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}$

The zeros of the denominator are.

  $\begin{array}{l}\left(x+3\right)\left(x-2\right)=0\\ x=-3,2\end{array}$

Clearly $\left(x+3\right)$ is a factor for both numerator and denominator.

Hence $x=-3$ is a hole and $x=2$ is a vertical asymptote.

Here degree of the numerator is two

And the degree of the denominator is two

Degree of the denominator is equal to the degree numerator.

Hence the horizontal asymptote is

  $\begin{array}{l}y=\frac{\text{leading}\text{coefficient}\text{in}\text{numerator}}{\text{leading}\text{coefficient}\text{in}\text{denominator}}\\ y=\frac{1}{1}\\ y=1\end{array}$

Hence the horizontal asymptote is $y=1$

Thus the rational function which contain a hole, horizontal asymptotes is

  $y=\frac{\left(x+6\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}$