Chapter 11.7, Problem 47PPE

To determine

To solve the function $y=\frac{2}{(x-2)(x+2)}$ with a dashed line.

Answer to Problem 47PPE

Dashed line $x=2$ and $x=-2$ are vertical asymptote

Explanation of Solution

Given:

Given function

  $y=\frac{2}{(x-2)(x+2)}$

Calculation:

In order to solve the equation $y=\frac{2}{(x-2)(x+2)}$ ,

  $\begin{array}{l}y=\frac{2}{(x-2)(x+2)}\\ here\\ x-2=0x+2=0\\ x=2x=-2\\ theyareverticalasymptote\end{array}$

Considering the values of x and y put the values in function

x	$y=\frac{2}{(x-2)(x+2)}$	(x, y)
-6	$y=\frac{2}{(x-2)(x+2)}$  y = 0.0625	(-6, 0.0625)
-4	$y=\frac{2}{(x-2)(x+2)}$  y = 0.166	(-4, 0.166)
-2	$y=\frac{2}{(x-2)(x+2)}$  y = 0	(-2, 0)
0	$y=\frac{2}{(x-2)(x+2)}$  y = -0.5	(0, -0.5)
2	$y=\frac{2}{(x-2)(x+2)}$  y = 0	(2, 0)
4	$y=\frac{2}{(x-2)(x+2)}$  y = 0.166	(4, 0.166)
6	$y=\frac{2}{(x-2)(x+2)}$  y = 0.0625	(6, 0.0625)

  

For a dashed line see denominator of function $x=2$ and $x=-2$

So the graph in both the quadrant 1^st and 2^nd translated by 2 unit left and right.

  $x=2$ and $x=-2$ are dashed line of vertical asymptote

Thus, the solution is correct.