Exercise 12 Page 526

We want to explain the relation between the given two functions. Notice that $y_2$ is the transformation of $y_1.$ To define this transformation, we will recall the possible translations.

Translations of $\mathbf{f}(\mathbf{x})=\frac{1}{\mathbf{x}},\mathbf{x}\ne 0$
Vertical Translations	$$\begin{array}{c}\text{Translation up }k\text{ units, }k>0\\ y=\frac{1}{x}+k\end{array}$$
	$$\begin{array}{c}\text{Translation down }k\text{ units, }k>0\\ y=\frac{1}{x}-k\end{array}$$
Horizontal Translations	$$\begin{array}{c}\text{Translation right }h\text{ units, }h>0\\ y=\frac{1}{x-h}\end{array}$$
	$$\begin{array}{c}\text{Translation left }h\text{ units, }h>0\\ y=\frac{1}{x+h}\end{array}$$

Using this table, let's indicate the translation from the graph of the parent function $f(x)=\frac{1}{x}$ to the graph of $y_1$ and $y_2.$

Function	Translation
$y_1=\frac{1}{x-1}+2$	Translation up $2$ units and right $1$ unit
$y_2=\frac{1}{x+1}-2$	Translation down $2$ units and left $1$ unit

Let's draw the graphs of these two functions on the coordinate plane.

Finally, let's consider a point on the graph of $y_1.$ We will see the translation applied to obtain its corresponding point on the graph of $y_2.$ For simplicity, let's use the point $(0,1).$

Therefore, the graph of $y_2$ is the graph of $y_1$ translated $2$ units to the left and $4$ units down.