Chapter 3.7, Problem 70E

To determine

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

Answer to Problem 70E

The quotient $y=\frac{x}{2}-\frac{1}{4}$ is the slant asymptote.

Vertical asymptotes are $x=-1,x=\frac{1}{2}$ .

Explanation of Solution

Given information:

Given function,

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

Calculation:

Consider the function,

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

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 $2x^2+x-1$ .

So,

  $2x^2+x-1=0$

On solving, we get,

  $\begin{array}{c}x=\frac{-1\pm \sqrt{1-4\left(2\right)\left(-1\right)}}{2\cdot 2}\\ =\frac{-1\pm \sqrt{1+8}}{2\cdot 2}\\ =\frac{-1\pm \sqrt{9}}{2\cdot 2}\\ =\frac{-1\pm 3}{2\cdot 2}\\ =\frac{-4}{4},\frac{2}{4}\\ =-1,\frac{1}{2}\end{array}$

Therefore, vertical asymptotes are $x=-1,x=\frac{1}{2}$ .

Using long division, we get

  $\begin{array}{c}2x^2+x-1\begin{array}{c}\hfill \frac{x}{2}-\frac{1}{4}\\ \hfill \overline{)x^3+4}\end{array}\\ \underset{\_}{x^3+\frac{x^2}{2}-\frac{x}{2}}\\ \text{}\text{}-\frac{x^2}{2}-\frac{x}{2}+4\\ \underset{\_}{-\frac{x^2}{2}-\frac{x}{4}+\frac{1}{4}}\\ \frac{3x}{4}+\frac{17}{4}\end{array}$

This implies that,

  $r\left(x\right)=\frac{x}{2}-\frac{1}{4}+\frac{\frac{3x+17}{4}}{2x^2+x-1}$

Therefore, the quotient $y=\frac{x}{2}-\frac{1}{4}$ is the slant asymptote.

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

Thus, the required graph is given below,