Chapter 2.7, Problem 92E

To determine

To graph: the function $g(x)=\frac{x^2+x-2}{x-1}$

Explanation of Solution

Graph:

First find some points that can be used to draw the graph

At $x=0,g(x)=\frac{0-2}{0-1}=2$

At $x=2,g(x)=\frac{4+2-2}{2-1}=\frac{4}{1}=4$

At $x=3,g(x)=\frac{9+3-2}{3-1}=\frac{10}{2}=5$

At $x=5,g(x)=\frac{25+5-2}{5-1}=\frac{28}{4}=7$

Plot the points $\left(0,2\right),\left(2,4\right),\left(3,5\right),\left(5,7\right)$

  

A rational function has generally vertical asymptotes at the points where the denominator is equal to $0$

Consider the function $g(x)=\frac{x^2+x-2}{x-1}$

The denominator is $x-1$ that is equal to $0$ at $x=1$

So, it appears that it can have vertical asymptote at $x=1$ .

But actually $g(x)=\frac{x^2+x-2}{x-1}=\frac{x^2+2x-x-2}{x-1}=\frac{x(x+2)-1(x+2)}{x-1}=\frac{(x-1)(x+2)}{x-1}=x+2$

Here, $x-1$ cancel out from both the numerator and denominator.

As a result, the function has no vertical asymptote at $x=1$