Chapter 13, Problem 54SGR

To determine

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

Answer to Problem 54SGR

Amplitude $=undefined$

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

Phase shift is ${30}^0$ right.

Vertical shift is $D=-3$ three units down

Explanation of Solution

Given information:

  $y=\frac{1}{2}\tan \left[2\left(\theta -{30}^0\right)\right]-3$

Concept:

If equation in the form

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

Then

Amplitude $=undefined$

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

Phase shift $=C$ .

Vertical shift $=D$

Given

  $y=\frac{1}{2}\tan \left[2\left(\theta -{30}^0\right)\right]-3$

On comparing with general equation

We get

  $\begin{array}{l}A=\frac{1}{2}\\ B=2\\ C={30}^0\\ D=-3\end{array}$

Graph of given equation is

  

X-axis is represented by $\theta$

Therefore,

Amplitude $=A=undefined$

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

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

Phase shift

Since $C={30}^0$

Phase shift is ${30}^0$ right.

Vertical shift is $D=-3$ three unitsdown

Therefore,

We got

Amplitude $=undefined$

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

Phase shift is ${30}^0$ right.

Vertical shift is $D=-3$ three unitsdown