Chapter 12.8, Problem 58PPS

To determine

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

Answer to Problem 58PPS

The period, phase shift and vertical shift of the function is $4\pi ,phaseshift:{90}^{\circ }totheright\&verticalshift:0$ .

Explanation of Solution

Given: It is given in the question that the function is $y=2\sec \frac{1}{2}(\theta -{90}^{\circ })$ .

Concept Used:

In this, use the concept of the general equation for secant $y=a\sec b(\theta -h)+d$ , $\frac{2\pi }{\left|b\right|}$ 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 and d is vertical shift.

Calculation: Here, the function is $y=2\sec \frac{1}{2}(\theta -{90}^{\circ })$ .

Now, the value of b is $\frac{1}{2}$ , so the period is: $\frac{2\pi }{\left|b\right|}=\frac{2\pi }{\left|\frac{1}{2}\right|}=4\pi$

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

Again, the vertical shift d is d units up if d is positive or d is down if d is negative.

So, in this case $d=0$ so the vertical shift is zero.

The graph of the function is given below:

  

Conclusion:

The period, phase shift and vertical shift are $4\pi ,phaseshift:{90}^{\circ }totheright\&verticalshift:0$ .