Problem 26 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=-3 \sin \left(2 x+\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The function \(y=-3 \sin \left(2 x+\frac{\pi}{2}\right)\) has an amplitude of 3, a period of \(\pi\), and is shifted right by \(\frac{\pi}{4}\).

Step 1: Finding the Amplitude

In the given function \(y=-3 \sin \left(2 x+\frac{\pi}{2}\right)\), the amplitude is represented by the absolute value of the coefficient before the sine function, therefore, the amplitude is \(|-3|\)=3.

Step 2: Calculating the Period

The period is calculated using the formula \(2\pi / |b|\), where \(b\) in our equation is the coefficient of \(x\) inside the sine function. So the period is \(2\pi / |2|\)= \(\pi\).

Step 3: Determining the Phase Shift

The phase shift is found by \(-c/b\), where \(c\) is the constant added to \(x\) inside the sine function and \(b\) is the coefficient of \(x\). It can be determined as \(-\frac{\pi /2}{2}= -\frac{\pi}{4}\). This means the graph is shifted to the right by \(\frac{\pi}{4}\).

Step 4: Graphing the Function

Finally, when creating the graph, keep in mind the amplitude (maximum and minimum values along y-axis), period (length of one cycle along x-axis), and phase shift (horizontal shift). The function begins and ends at -3 since it is negative, finishes one period at \(\pi\), and is shifted to the right by \(\frac{\pi}{4}\).