Chapter 12, Problem 53SGR

To determine

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

Answer to Problem 53SGR

Amplitude $=A=3$

Period $=\pi$

Phase shift is ${90}^0$ right.

Vertical shift is $D=1$ one unit up

Explanation of Solution

Given information:

  $y=3\sin \left[2\left(\theta -{90}^0\right)\right]+1$

Concept:

If equation in the form

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

Then

Amplitude $=A$

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

Phase shift $=C$ .

Vertical shift $=D$

Given

  $y=3\sin \left[2\left(\theta -{90}^0\right)\right]+1$

On comparing with general equation

We get

  $\begin{array}{l}A=3\\ B=2\\ C={90}^0\\ D=1\end{array}$

Graph of given equation is

  

Therefore,

Amplitude $=A=3$

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

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

  $\Rightarrow$ Period $=\pi$

Phase shift

Since $C={90}^0$

Phase shift is ${90}^0$ right.

Vertical shift is $D=1$ one unit up

Therefore,

We got

Amplitude $=A=3$

Period $=\pi$

Phase shift is ${90}^0$ right.

Vertical shift is $D=1$ one unit up