Chapter 2.7, Problem 49E

To determine

To sketch the function $h(x)=\frac{x^2}{x-1}$ , also find their intercepts and asymptotes.

Answer to Problem 49E

  $x=1,y=x+1$

Explanation of Solution

Given:

Function: $h(x)=\frac{x^2}{x-1}$

Calculation for graph:

Consider $h(x)=\frac{x^2}{x-1}$

Values of x	Values of h(x)
0	0
1	Infinity
-1	-0.5
2	4
-2	-1.333

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

Graph:

  

Calculation:

Intercepts:

Let $h(x)=y.$

  $\Rightarrow y=\frac{x^2}{x-1}$

To find x intercepts, put y = 0,

  $\begin{array}{l}\Rightarrow y=0\\ \Rightarrow \frac{x^2}{x-1}=0\\ \Rightarrow x^2=0\\ \Rightarrow x=0\end{array}$

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

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{{(0)}^2}{0-1}\\ \Rightarrow y=\frac{0}{-1}\\ \Rightarrow y=0\end{array}$

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

Asymptotes:

Vertical asymptotes:

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

  $\begin{array}{l}\Rightarrow x-1=0\\ \Rightarrow x=1\end{array}$

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: $x^2$

Divisor: $x-1$

  $x-1\begin{array}{c}\hfill x+1\\ \hfill \overline{)\begin{array}{l}{x}^2+0x+0\\ \begin{array}{c}{x}^2-x\end{array}\\ x+0\\ \begin{array}{c}x-1\end{array}\\ 1\end{array}}\end{array}$  $\begin{array}{l}\frac{x^2}{x}=x\\ \text{Multiply}x(x-1)\text{and subtract}\text{.}\\ \\ \frac{x}{x}=1\\ \text{Multiply}1(x-1)\text{and subtract}\text{.}\end{array}$

Here, quotient is $x+1$ and remainder is 1.

So, the slant asymptote is $y=x+1.$