Chapter 13.8, Problem 15PPS

To determine

Find the amplitude, period and phase shift for each function. Also, draw the graph of the function.

Answer to Problem 15PPS

The amplitude, period and phase shift of the function is $doesnotexist,{180}^{\circ },phaseshift{90}^{\circ }totheright$ .

Explanation of Solution

Given: It is given in the question that the function is $y=\tan (\theta -{90}^{\circ })$ .

Concept Used:

In this, use the concept of the general equation for tangent $y=a\tan b(\theta -h)$ , where a is amplitude which does not exist, $\frac{\pi }{\left|b\right|}$ is period and h is the phase shift and if the h is positive , then h units to the right and if h is negative then h units to the left.

Calculation: Here, the function is $y=\tan (\theta -{90}^{\circ })$ .

In this, $a=1$ , but it is a tangent function, so the amplitude does not exist.

Now, the value of b is $1$ , so the period is: $\frac{\pi }{\left|b\right|}=\frac{180}{\left|1\right|}=180$

From the equation it is find that h is ${90}^{\circ }$ with is positive sign, so the phase shift is ${90}^{\circ }$ to the right.

The graph of the function is given below:

Conclusion:

The amplitude, period and phase shift are $doesnotexist,{180}^{\circ },phaseshift{90}^{\circ }totheright$ .