Chapter 4, Problem 120RE

To determine

To find: the period and amplitude.

Answer to Problem 120RE

Amplitude $=\frac{3}{2}=1.5$

Period $=4\pi$

Explanation of Solution

Given:

  

Calculation:

The given graph is of the function $y=-\frac{3}{2}\sin \left(\frac{x}{2}\right)$

For the functions $y=A\sin Bx$ and $y=A\cos Bx$

Amplitude is given by $\left|A\right|$

And the period is given by $\frac{2\pi }{B}$

Now, for the given function,

  $A=-\frac{3}{2},B=\frac{1}{2}$

Hence, the amplitude is $\left|-\frac{3}{2}\right|=\frac{3}{2}=1.5$

And the period is given by

  $\begin{array}{l}\frac{2\pi }{B}\\ =\frac{2\pi }{\frac{1}{2}}\\ =4\pi \end{array}$