Chapter 9.4, Problem 22PPS

To determine

To graphthe given function.

Explanation of Solution

Given information:

The function is given as,

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

Graph:

Find the zeros,

  $\begin{array}{l}4x^3=0\text{}\left[\text{set }a\left(x\right)=0\right]\\ x^3=0\\ x=0\end{array}$

There is a zero at $x=0$ .

Find the asymptotes,

  $\begin{array}{l}2x^2+x-1=0\text{}\left[\text{set }b\left(x\right)=0\right]\\ \left(2x-1\right)\left(x+1\right)=0\\ 2x-1=0\text{ or }x+1=0\\ x=\frac{1}{2}\text{ or }x=-1\end{array}$

There is a vertical asymptote at $x=-1\text{ and }x=\frac{1}{2}$ .

The degree of numerator is greater than the degree of the denominator, so there is no horizontal asymptote.

The difference between the degree of numerator and the degree is 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 asymptote is the quotient excluding any remainder.

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

Thus, the oblique asymptote is the line, $f\left(x\right)=2x-1$

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

$x$	$y$
$-4$	$-9.5$
$-2$	$-6.4$
0	$0$
2	$3.56$
4	$7.3$

The graph is obtained as,

  

Interpretation:

The graph of the given function is obtained with a vertical asymptote at $x=-1\text{ and }x=\frac{1}{2}$ , oblique asymptote is the line $f\left(x\right)=2x-1$ .