Chapter 3.7, Problem 71E

To determine

To find: the slant asymptote, the vertical asymptotes, and sketch a graph of the function.

Answer to Problem 71E

The quotient $y=x+1$ is the slant asymptote.

Vertical asymptotes are $x=\pm 2$ .

Explanation of Solution

Given information:

Given function,

  $r\left(x\right)=\frac{x^3+x^2}{x^2-4}$

Calculation:

Consider the function,

  $r\left(x\right)=\frac{x^3+x^2}{x^2-4}$

Vertical asymptote of $r$ are the lines $x=a$ , where $a$ is a zero of the denominator.

Al slant asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator. To find the slant asymptote must divide the numerator by the denominator using either long division or synthetic division.

Here, denominator is $x^2-4$ .

So,

  $x^2-4=0$

On solving, we get,

  $\begin{array}{c}{x}^2-4=0\\ x^2=4\\ x=\pm 2\end{array}$

Therefore, vertical asymptotes are $x=\pm 2$ .

Using long division, we get

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

This implies that,

  $r\left(x\right)=x+1+\frac{4x-4}{x^2-4}$

Therefore, the quotient $y=x+1$ is the slant asymptote.

Now to draw the graph of the function $r\left(x\right)=\frac{x^3+x^2}{x^2-4}$ , slant asymptote $y=x+1$ and vertical asymptotes $x=2,x=-2$ .

Thus, the required graph is given below,