Chapter 9.4, Problem 20PPS

To determine

To graph: The given function.

Explanation of Solution

Given information:

The following function:

  $f\left(x\right)=\frac{{(x+3)}^2}{x-5}$

Calculation:

Step 1:

Find the zeroes.

  $\begin{array}{l}$ {(x+3)}^2=0[\text{Set }a(x)=0]$x+3=0[\text{Take square root of each side}]\\ x=-3[\text{Subtract 3 from each side}]\end{array}$

There is a zero at $x=-3$ .

Step 2:

Find the asymptotes.

  $\begin{array}{l}x-5=0[\text{Set }b(x)=0]\\ x=5[\text{Add 5 to each side]}\end{array}$

There is a vertical asymptote at $x=5$ .

The degree of numerator is greater than the degree of the denominator. Therefore, there is no horizontal asymptote.

The difference between the degree of the numerator and the degree of the denominator is $1$ , so there is an oblique asymptote.

Divide the numerator by the denominator to determine the equation of the oblique asymptote.

  $\begin{array}{l}x-5\begin{array}{c}\hfill x+11\\ \hfill \overline{)x^2+6x+9}\end{array}\text{ [Expand the numerator: (}x+3{)}^2=x^2+6x+9]\\ \underset{\_}{(-)x^2(+)-5x}\\ 11x+9\\ \underset{\_}{\text{(}-\text{)11}x(+)-55}\\ \text{ 64}\end{array}$

The equation of the asymptote is the quotient excluding any remainder.

Thus, the oblique asymptote is the line

  $f(x)=x+11$

Step 3:

Draw the graph

Use a table to find ordered pairs on the graph. Then connect the points.

The table of values:

  

The graph: