Chapter 8.4, Problem 3BGP

To determine

To graph: The given function.

Explanation of Solution

Given information:

The following function:

  $f\left(x\right)=\frac{x^3-1}{x^2-4}$ .

Calculation:

Step 1:

Find the zeroes.

  $\begin{array}{l}{x}^3-1=0[\text{Set }a(x)=0]\\ x^3=1[\text{Add 1 to both sides]}\\ x=1[\text{Find cube roots of both sides]}\end{array}$

There is a zero at $x=1$ .

Step 2:

Find the asymptotes.

  $\begin{array}{l}{x}^2-4=0[\text{Set }b(x)=0]\\ \left(x+2\right)\left(x-2\right)=0[\text{Factor]}\\ x+2=0x-2=0[\text{Set each factor to 0]}\\ x=-2x=2[\text{Simplify]}\end{array}$

There is a vertical asymptote at $x=2$ and $x=-2$ .

The degree of numerator is greater than the degree of the denominator.

So, 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.

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

  $x^2-4\begin{array}{c}\hfill x\\ \hfill \overline{)\begin{array}{l}{x}^3-1\\ \underset{\_}{(-)x^3-4x}\\ \text{ 4}x-1\\ \end{array}}\end{array}$

Thus, the oblique asymptote is the line $f(x)=x$ .

Step 3:

Draw the asymptotes, and then use a table of values to graph the function.