Problem 15 - Precalculus Essentials

Question

Determine the amplitude and period of each function. Then graph one period of the function. $$y=-\sin \frac{2}{3} x$$

Step-by-Step Solution

Answer

The amplitude of the given function is 1 and the period is \(3\pi\). The function starts from zero, hits the maximum or minimum value halfway through the period, and returns to zero at the end of the period.

Step 1: Determine the Amplitude

Look at the coefficient in front of the sine function, which is -1. The amplitude of a sine or cosine function is the absolute value of that coefficient, so in this case, \(|\(-1\)| = 1. So, the amplitude is 1.

Step 2: Determine the Period

The standard period of a sine function is \(2\pi\). However, if the x is multiplied by a factor, in this case, \(\frac{2}{3}\), the period changes. The period is calculated by dividing \(2\pi\) by the absolute value of the coefficient of x, \(\frac{2}{3}\). So, the new period is \(\frac{2\pi}{\(\frac{2}{3}\)} = 3\pi\).

Step 3: Plot the function

To draw a complete period of the function, \(y=-\sin \frac{2}{3}x\), start at x=0. Due to the negative sign in front of the sine function, the function starts at 0, then goes down to -1 at \(x=\frac{3\pi}{4}\), comes back to 0 at \(x=\frac{3\pi}{2}\), goes to 1 at \(x=\frac{9\pi}{4}\), and comes back to 0 at \(x=3\pi\). It continues in this way for subsequent periods. The wave goes both above and below the x-axis at 1 and -1, respectively, adhering to our calculated amplitude.