Chapter 13.8, Problem 29PPS

To determine

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

Answer to Problem 29PPS

The amplitude, period , phase shift and vertical shift of the function is $doesnotexist,\frac{\pi }{2},phaseshift\frac{\pi }{4}totheleft\&\text{vertical shifts}:5unit\mathrm{s}down.$

Explanation of Solution

Given: It is given in the question that the function is $y=2\tan 2(\theta +\frac{\pi }{4})-5$ .

Concept Used:

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

Calculation: Here, the function is $y=2\tan 2(\theta +\frac{\pi }{4})-5$ .

In this, $a=1$ , but it is a tangent function so the amplitude doesn’t exist.

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

The phase shift of the function $y=a\tan b(\theta -h)+d$ is h units to the right if h is positive or h units to the left if h is negative.

In this case, $h=-\frac{\pi }{4}$ so the phase shift is $\frac{\pi }{4}$ to the left.

From the equation it is seen that $d=-5$ so the vertical shifts is five units down.

The graph of the function is given below:

  

Conclusion:

The amplitude, period , phase shift and vertical shift are $doesnotexist,\frac{\pi }{2},phaseshift\frac{\pi }{4}totheleft\&\text{vertical shifts}:5unit\mathrm{s}down.$