Chapter 3.7, Problem 92E

a)

To determine

To graph : The rational function, $r\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ using the graph transformations on $y=\frac{1}{x^2}$ .

Explanation of Solution

Given information : The rational function is $r\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ .

Graph : Draw the graph of $y=\frac{1}{x^2}$ .

  

Shift the graph of $y=\frac{1}{x^2}$ horizontally to the right by 2 units to make the graph of $r\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ .

  

Interpretation : The graph of $y=\frac{1}{x^2}$ is shifted horizontally by 2 units to draw the graph of $r\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ .

b)

To determine

To prove : The function, $s\left(x\right)=\frac{2x^2+4x+5}{x^2+2x+1}$ can be written as $s\left(x\right)=2+\frac{3}{{\left(x+1\right)}^2}$ using long division method. Also, graph the function, $s\left(x\right)$ using the graph transformations on $y=\frac{1}{x^2}$ .

Explanation of Solution

Given information : The function is $s\left(x\right)=\frac{2x^2+4x+5}{x^2+2x+1}$ .

Proof : Use long division to divide numerator with denominator.

  $\begin{array}{l}{x}^2+2x+1\begin{array}{c}\hfill 2\\ \hfill \overline{)2x^2+4x+5}\end{array}\\ \underset{\_}{(-)2x^2+4x+2}\\ 3\end{array}$

The function can be expressed as,

  $\begin{array}{c}s\left(x\right)=2+\frac{3}{x^2+2x+1}\\ =2+\frac{3}{x^2+2\cdot 1\cdot x+1^2}\\ =2+\frac{3}{{\left(x+1\right)}^2}\end{array}$

Thus, the function, $s\left(x\right)=\frac{2x^2+4x+5}{x^2+2x+1}$ can be written as $s\left(x\right)=2+\frac{3}{{\left(x+1\right)}^2}$ .

Draw the graph of $y=\frac{1}{x^2}$ .

  

Shift the graph of $y=\frac{1}{x^2}$ horizontally to the left by 1 units to make the graph of $t\left(x\right)=\frac{1}{{\left(x+1\right)}^2}$ .

Stretch the graph of $t\left(x\right)=\frac{1}{{\left(x+1\right)}^2}$ vertically by 3 units to draw the graph of $u\left(x\right)=\frac{3}{{\left(x+1\right)}^2}$ .

Shift the graph of $u\left(x\right)=\frac{3}{{\left(x+1\right)}^2}$ vertically upwards by 2 units to make the graph of $s\left(x\right)=2+\frac{3}{{\left(x+1\right)}^2}$ .