Chapter 4.4, Problem 45E

Solution

To state: The transformation required to obtain the graph of $y=-\frac{2}{3}\cos \frac{x}{3}$ from a basic trigonometric graph.

The graph will stretch horizontally by $3$ , shrink vertically by $\frac{2}{3}$ and reflected across x- axis from the graph of $y=\cos x$ .

Given information:

The given function is $y=-\frac{2}{3}\cos \frac{x}{3}$ .

Explanation:

Consider the below given information:

If $\left|b\right|>1$ then there is a horizontal shrink by the factor of $\frac{1}{\left|b\right|}$ .

If $\left|b\right|<1$ then there is a horizontal stretch by the factor of $\frac{1}{\left|b\right|}$ .

If $\left|b\right|<0$ then there is a reflection across the y- axis.

Now consider the given function: $y=-\frac{2}{3}\cos \frac{x}{3}$

Compare the function $y=-\frac{2}{3}\cos \frac{x}{3}$ with the standard function $y=a\cos bx$ .

Then the values are $a=-\frac{2}{3}\text{ and }b=\frac{1}{5}$ .

Here $\left|a\right|$ is the amplitude of the function and its value is $\frac{2}{3}$ .

Thus the graph of $y=-\frac{2}{3}\cos \frac{x}{3}$ is vertical shrink of graph of $y=\cos x$ by a factor of $\frac{2}{3}$ and reflected across x- axis.

Now find the period of the function.

  $\text{Period}=\frac{2\pi }{\left|b\right|}$

Substitute the values $b=\frac{1}{3}$ in the formula:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{\left|\frac{1}{3}\right|}\\ =6\pi \end{array}$

Thus the graph of $y=-\frac{2}{3}\cos \frac{x}{3}$ is vertical shrink of graph of $y=\cos x$ by a factor of $\frac{2}{3}$ and horizontal stretch by a factor of 3 and reflected across x- axis.