Chapter 13.8, Problem 9CYU

To determine

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

Answer to Problem 9CYU

The amplitude, period , phase shift and vertical shift of the function is $2,{360}^{\circ },phaseshift{145}^{\circ }\text{vertical shifts}:1unit\mathrm{s}up.$

Explanation of Solution

Given: It is given in the question that the function is $y=2\sin (\theta +{45}^{\circ })+1$ .

Concept Used:

In this, use the concept of the general equation for sine $y=a\sin b(\theta -h)+d$ , where a is amplitude which does not exist, $\frac{2\pi }{\left|b\right|}$ is period and d is the vertical shift and h is the phase shift.

Calculation: Here, the function is $y=2\sin (\theta +{45}^{\circ })+1$ .

In this, $a=2$ ,so the amplitude is $2$ .

Now, the value of b is $1$ , so the period is: $\frac{2\pi }{\left|b\right|}=\frac{360}{\left|1\right|}={360}^{\circ }$ .

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

In this case, $h=-{45}^{\circ }$ so the phase shift is ${45}^{\circ }$ to the left.

From the equation it is seen that $d=1$ so the vertical shifts is one units up.

The graph of the function is given below:

  

Conclusion:

The amplitude, period , phase shift and vertical shift are $2,{360}^{\circ },phaseshift{145}^{\circ }\text{vertical shifts}:1unit\mathrm{s}up.$