Chapter 3.7, Problem 66E

To determine

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

Answer to Problem 66E

The quotient $y=x+3$ is the slant asymptote.

Vertical asymptote is $x=1$ .

Explanation of Solution

Given information:

Given function,

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

Calculation:

Consider the function,

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

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 $x-1$ .

So,

  $x-1=0$

This implies that,

  $x=1$

Therefore, vertical asymptote is $x=1$ .

Using long division, we get

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

This implies that,

  $r\left(x\right)=\left(x+3\right)+\frac{3}{x-1}$

Therefore, the quotient $y=x+3$ is the slant asymptote.

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

Thus, the required graph is given below,