Chapter 2.7, Problem 51E

To determine

To sketch the function $g(x)=\frac{x^3}{2x^2-8}$ , also find their intercepts and asymptotes.

Answer to Problem 51E

  $x=2,-2,y=\frac{x}{2}$

Explanation of Solution

Given:

Function: $g(x)=\frac{x^3}{2x^2-8}$

Calculation for graph:

Consider $g(x)=\frac{x^3}{2x^2-8}$

Values of x	Values of g(x)
0	0
1	-0.166
-1	0.166
2	Infinity
-2	Infinity

By taking different values of x, the graph can be plotted.

Graph:

  

Calculation:

Intercepts:

Let $g(x)=y.$

  $\Rightarrow y=\frac{x^3}{2x^2-8}$

To find x intercepts, put y = 0,

  $\begin{array}{l}\Rightarrow y=0\\ \Rightarrow \frac{x^3}{2x^2-8}=0\\ \Rightarrow x^3=0\\ \Rightarrow x=0\end{array}$

So, the x intercept is (0, 0).

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{{(0)}^3}{2{(0)}^2-8}\\ \Rightarrow y=\frac{0}{-8}\\ \Rightarrow y=0\end{array}$

So, the y intercept is (0, 0).

Asymptotes:

Vertical asymptotes:

To find vertical asymptotes, put denominator of the given function equal to zero.

  $\begin{array}{l}\Rightarrow 2x^2-8=0\\ \Rightarrow 2x^2=8\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm \sqrt{4}\\ \Rightarrow x=\pm 2\end{array}$

Horizontal asymptotes:

As the degree of numerator is larger than the degree of the denominator, the horizontal asymptote does not exist for given function.

Slant asymptotes:

To find slant asymptote, divide given function numerator with denominator using long division.

Dividend: $x^3$

Divisor: $2x^2-8$

  $2x^2-8\begin{array}{c}\hfill \frac{x}{2}\\ \hfill \overline{)\begin{array}{l}{x}^3+0x^2+0x+0\\ \begin{array}{c}{x}^3+0x^2-4x\end{array}\\ 4x\end{array}}\end{array}$  $\begin{array}{l}\frac{x^3}{2x^2}=\frac{x}{2}\\ \text{Multiply}x(2x^2-8)\text{and subtract}\text{.}\end{array}$

Here, quotient is $\frac{x}{2}$ and remainder is 4x.

So, the slant asymptote is $y=\frac{x}{2}.$