Chapter 13.8, Problem 6CYU

To determine

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

Answer to Problem 6CYU

The amplitude, period and vertical shift and the equation of the midline of the function is $1,{360}^{\circ },\text{vertical shifts}:2unit\mathrm{s}down,y=-2$

Explanation of Solution

Given: It is given in the question that the function is $y=\sin \theta -2$ .

Concept Used:

In this, use the concept of the general equation for sine $y=a\sin b\theta +d$ , where a is amplitude, $\frac{2\pi }{\left|b\right|}$ is period and d is the vertical shift and y = d is midline.

Calculation: Here, the function is $y=\sin \theta -2$ .

In this, $a=1$ , so the amplitude is $1$ .

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

From the equation it is seen that $d=-2$ so the vertical shifts is two units down.

Also, $d=-2$ then the midline is $y=-2$ .

The graph of the function is given below:

  

Conclusion:

The amplitude, period and vertical shift and the equation of midline are $1,{360}^{\circ },\text{vertical shifts}:2unit\mathrm{s}down,y=-2$ .