Chapter 2.7, Problem 26E

To determine

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

Answer to Problem 26E

  $x=0,3,y=0$

Explanation of Solution

Given:

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

Calculation for graph:

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

Values of x	Values of h(x)
0	Infinity
1	-1
-1	-0.5
2	-0.5
-2	-0.1

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

Graph:

  

Calculation:

Intercepts:

Let $h(x)=y.$

  $\Rightarrow y=\frac{2}{x^2(x-3)}$

To find x intercepts, put y = 0,

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

Here, the answer is wrong. It means that, the given function does not have x intercepts.

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{2}{0(0-3)}\\ \Rightarrow y=\frac{2}{0}\to \text{Undefined}\end{array}$

So, y intercept does not exist for given function.

Asymptotes:

Vertical asymptotes:

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

  $\begin{array}{l}\Rightarrow x^2(x-3)=0\\ \\ \begin{array}{cc}\begin{array}{l}\Rightarrow x^2=0\\ \Rightarrow x=0\end{array}& \begin{array}{l}\Rightarrow x-3=0\\ \Rightarrow x=3\end{array}\end{array}\\ \\ \Rightarrow x=0,3\end{array}$

Horizontal asymptotes:

As the degree of numerator is less than the degree of the denominator, the horizontal asymptote is $y=0.$

Holes:

Here, the given function does not contain common factor in numerator and denominator. So, the function does not have any holes.