Chapter 2, Problem 16CT

To determine

To find: Intercepts and asymptotes of the given function as well as plot graph.

Answer to Problem 16CT

There are no x-intercepts and no y-intercepts.

There are no horizontal asymptotes of the graph of the function

The vertically asymptotes of function is $x=1$

Explanation of Solution

Given:

The function:

  $g\left(x\right)=\frac{x^2+2}{x-1}$

Consider the following function

  $g\left(x\right)=\frac{x^2+2}{x-1}$

To find the x-intercepts put $g\left(x\right)=0$ and solve for x

  $\begin{array}{l}g\left(x\right)=0\\ \frac{x^2+2}{x-1}=0\\ x^2+2=0\end{array}$

Since the above equation has no real roots, therefore there are no x-intercepts

To find the y-intercept, put $x=0$

  $\begin{array}{l}g(x)=\frac{{\left(0\right)}^2+2}{0-1}\\ =\frac{2}{-1}\\ =-2\end{array}$

Thus, the function has no y-intercepts.

From the above function, note that as x tends to infinity, the function also approaches to infinity

Thus, there are no horizontal asymptotes of the graph of the function

To find the vertically asymptotes, put the denominator of the function to zero and solve for x

  $\begin{array}{l}x–1=0\hfill \\ x=0\hfill \end{array}$

Thus, the vertically asymptotes of function is $x=1$

Consider the table

Test interval	x	y	Sign	Point of graph
$\left(–\infty ,–1\right)$	−3	$–\frac{11}{4}$	Negative	$\left(–3,–\frac{11}{4}\right)$
$\left(1,\infty \right)$	2	6	Positive	(2,6)
$\left(1,\infty \right)$	3	$\frac{11}{2}$	Positive	$\left(3,\frac{11}{2}\right)$

Thus, the graph of the function is shown below