Chapter 14.7, Problem 19WE

To determine

To fill : in the blank.

Answer to Problem 19WE

If $T:\left(x,y\right)\to \left(x+3,y-4\right)$ then $T^2:\left(x,y\right)\to$

  $\underset{\_}{\left(x+6,y-8\right)}$ .

Explanation of Solution

Given information : If $T:\left(x,y\right)\to \left(x+3,y-4\right)$ then $T^2:\left(x,y\right)\to$

  $\underset{\_}{}$ .

If $T:\left(x,y\right)\to \left(x+3,y-4\right)$ then $T^2:\left(x,y\right)\to$

  $\underset{\_}{\left(x+6,y-8\right)}$ .

  $\begin{array}{l}T\left(x,y\right)=\left(x+3,y-4\right)\\ T^2\left(x,y\right)=T\left(T\left(x,y\right)\right)\\ T\left(x+3,y-4\right)\\ \Rightarrow \left(x+3+3,y-4-4\right)\\ \Rightarrow \left(x+6,y-8\right)\end{array}$