Chapter 6.5, Problem 27E

To determine

To state: The amplitude, period, phase shift and vertical shift of the sine curve given below.

Answer to Problem 27E

  $\begin{array}{l}\text{Amplitude: 4}\\ \text{Period: 2}\pi \\ \text{Phase shift: }\pi /2\text{ to the left}\\ \text{Vertical shift: }-1\end{array}$

Explanation of Solution

Given information: The curve:

Calculation:

Sine graphs starts at zero and go up, so there is a phase shift of $\pi /2$ to the left.

The difference from the max to the min point is 8, so the amplitude is 4.

The midpoint is at − 1, so there is a vertical shift of − 1 and the period is $2\pi$ .

Thus,

  $\begin{array}{l}\text{Amplitude: 4}\\ \text{Period: 2}\pi \\ \text{Phase shift: }\pi /2\text{ to the left}\\ \text{Vertical shift: }-1\end{array}$