Chapter 5.3, Problem 15E

Solution

To graph : The given function.

Given information :

Function $y=\frac{2x}{x^2-1}$

Graph :

  

Interpretation :

The numerator is zero when $x=0$ , so the $x\text{and}y$ -intercept is at $(0,0)$ .

As $x\to +\infty$ the denominator approaches $+\infty$ while the numerator is positive so $y$ approaches zero from above the $x$ axis and there is a horizontal asymptote is $y=0$ .

As $x\to -\infty$ the denominator approaches $+\infty$ while the numerator is negative so $y$ approaches zero from below the $x$ axis and the same horizontal asymptote is $y=0$ .

The zeros of the denominator are the vertical asymptotes.

  $\begin{array}{c}0=x^2-1\\ 0=(x+1)(x-2)\\ x=-1,1\end{array}$

Calculate some points, especially some close to the vertical asymptotes to see which way the graph goes.

$x$	$f(x)$
$-5.0$	$-0.42$
$-4.0$	$-0.53$
$-3.0$	$-0.75$
$-2.0$	$-1.33$
$-1.5$	$-2.40$
$-1.4$	$-2.92$
$-1.3$	$-3.77$
$-1.2$	$-5.45$
$-1.1$	$-10.48$
$-0.9$	$9.47$
$-0.8$	$4.44$
$-0.7$	$2.75$
$-0.6$	$1.87$
$0.0$	$0.00$
$0.5$	$-1.33$
$0.6$	$-1.88$
$0.7$	$-2.75$
$0.8$	$-4.44$
$0.9$	$-9.47$
$1.1$	$10.48$
$1.2$	$5.45$
$1.3$	$3.77$
$1.4$	$2.92$
$1.5$	$2.40$
$2.0$	$1.33$
$3.0$	$0.75$
$4.0$	$0.53$
$5.0$	$0.42$