Chapter 4.4, Problem 47E

Solution

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

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

Given information:

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

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: $y=3\cos \frac{2\pi x}{3}$

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

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

Here $a$ is the amplitude of the function and its value is 3.

Thus the graph of $y=3\cos \frac{2\pi x}{3}$ is vertical stretch of graph of $y=\cos x$ by a factor of $3$ .

Now find the period of the function.

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

Substitute the values $b=\frac{2\pi }{3}$ in the formula:

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

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