Chapter 3.3, Problem 52E

Solution

To describe : The transformation of the graph of $y=\log x$ into the graph of the given function.

The graph of $y=\log x$ is reflected across $y-$ axis followed by a translation of $1$ unit to the right, the vertically stretched by a factor of $-3$ and finally translated $1$ unit up.

Given information :

The function $f\left(x\right)=-3\log \left(1-x\right)+1$ .

Formula used:

The translation of a function $y=f\left(x\right)$ into the form $y=f\left(x\pm k\right)$ , represents the horizontal translation. If $k>0$ , then the function moves to the left side by $\text{'}k\text{'}$ units and if $k<0$ , then the function moves to the right by $\text{'}k\text{'}$ units.

The translation of a function $y=f\left(x\right)$ into the form $y=f\left(x\right)\pm k$ , represents the vertical translation. If $k>0$ , then the function moves up by $\text{'}k\text{'}$ units and if $k<0$ , then the function moves down by $\text{'}k\text{'}$ units.

  $y=-f\left(x\right)$ is the reflection of $y=f\left(x\right)$ with respect to the $x-$ axis.

  $y=f\left(-x\right)$ is the reflection of $y=f\left(x\right)$ with respect to the $y-$ axis.

Explanation :

Reflect the graph of $y=\log x$ across $y-$ axis followed by a translation of $1$ unit to the right to get $\log \left(1-x\right)$ .

Vertically stretch the graph of $\log \left(1-x\right)$ by a factor of $-3$ to get $-3\log \left(1-x\right)$ .

Translate the graph of $-3\log \left(1-x\right)$ by $1$ unit up to get $-3\log \left(1-x\right)+1$ .

The graph will be: