Chapter 12, Problem 56SGR

To determine

To find: Amplitude, Vertical shift, Phase shift and period of each function.

Answer to Problem 56SGR

Amplitude $=A=\frac{1}{2}$

Period $=8\pi$

Phase shift is $\frac{\pi }{4}$ units Left.

Vertical shift is $D=-1$ One unit Down

Explanation of Solution

Given information:

  $y=\frac{1}{2}\cos \left[\frac{1}{4}\left(\theta +\frac{\pi }{4}\right)\right]-1$

Concept:

If equation in the form

  $y=A\cos \left(B\left(x-C\right)\right)+D$

Then

Amplitude $=A$

Period $=\frac{2\pi }{B}$

Phase shift $=C$ .

Vertical shift $=D$

Given

  $y=\frac{1}{2}\cos \left[\frac{1}{4}\left(\theta +\frac{\pi }{4}\right)\right]-1$

On comparing with general equation

We get

  $\begin{array}{l}A=\frac{1}{2}\\ B=\frac{1}{4}\\ C=-\frac{\pi }{4}\\ D=-1\end{array}$

Graph of given equation is

  

X-axis is represented by $\theta$

Therefore,

Amplitude $=A=\frac{1}{2}$

Period $=\frac{2\pi }{B}$

  $\Rightarrow$ Period $=\frac{2\pi }{\frac{1}{4}}$

  $\Rightarrow$ Period $=2\pi \times 4$

  $\Rightarrow$ Period $=8\pi$

Phase shift

Since $C=-\frac{\pi }{4}$

Phase shift is $\frac{\pi }{4}$ units Left.

Vertical shift is $D=-1$ One unit Down

Therefore,

We got

Amplitude $=A=\frac{1}{2}$

Period $=8\pi$

Phase shift is $\frac{\pi }{4}$ units Left.

Vertical shift is $D=-1$ One unit Down