Chapter 8.3, Problem 56PPS

To determine

To Graph: The function.

Answer to Problem 56PPS

  

Explanation of Solution

Given: $f(x)=-\frac{1}{3}{\log }_6(6x+2)-5$

Given: $f(x)=-\frac{1}{3}{\log }_6(6x+2)-5$

Transform the given function

  $\begin{array}{l}f(x)=-\frac{1}{3}{\log }_6(6x+2)-5\\ =-\frac{1}{3}{\log }_6\left(6\left(x+\frac{2}{6}\right)\right)-5\\ =-\frac{1}{3}{\log }_{6}6+{\log }_6\left(x+\frac{1}{3}\right)-5\\ =-\frac{1}{3}-\frac{1}{3}{\log }_6\left(x+\frac{1}{3}\right)-5\\ =-\frac{1}{3}{\log }_6\left(x+\frac{1}{3}\right)-\frac{16}{3}\end{array}$

Thus

  $f(x)=-\frac{1}{3}{\log }_6\left(x+\frac{1}{3}\right)-\frac{16}{3}$

The graph of the function is

  

This shows the transformation of the function ${\log }_{6}x$

  $\left|a\right|=\left|-\frac{1}{3}\right|=\frac{1}{3}<1$ : The graph is compressed vertically

  $h=-\frac{1}{3}$ : The graph is translated $\frac{1}{3}$ units to the left.

  $k=-\frac{16}{3}$ : The graph is translated $\frac{16}{3}$ units to the down.