Chapter 3.7, Problem 68E

To determine

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

Answer to Problem 68E

The quotient $y=1-\frac{x}{2}$ is the slant asymptote.

Vertical asymptote is $x=1$ .

Explanation of Solution

Given information:

Given function,

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

Calculation:

Consider the function,

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

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 $2x-2$ .

So,

  $2x-2=0$

This implies that,

  $\begin{array}{l}2x=2\\ x=1\end{array}$

Therefore, vertical asymptote is $x=1$ .

Using long division, we get

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

This implies that,

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

Therefore, the quotient $y=1-\frac{x}{2}$ is the slant asymptote.

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

Thus, the required graph is given below,