Problem 10 - Precalculus Essentials

Question

Determine the amplitude and period of each function. Then graph one period of the function. $$y=2 \sin \frac{1}{4} x$$

Step-by-Step Solution

Answer

The amplitude of the function \(y = 2\sin\left(\frac{1}{4}x\right)\) is 2 and the period is \(8\pi\). In the graph, the function starts at y=0 at x=0, reaches 2 at \(2\pi\), returns to 0 at \(4\pi\), goes to -2 at \(6\pi\), and finally back to 0 at \(8\pi\).

Step 1: Determine the Amplitude

The amplitude of the function is typically the coefficient in front of the sine function. So here, the amplitude of the function \(y = 2\sin\left(\frac{1}{4}x\right)\) is 2.

Step 2: Determine the Period

The period of the sine function is given by \(2\pi/B\). In this case, B equals to 1/4, so the calculation gives \(2\pi/\frac{1}{4} = 8\pi\). So, the period of the function \(y = 2\sin\left(\frac{1}{4}x\right)\) is \(8\pi\).

Step 3: Graph the Function

To graph the function, first set up the x-axis from 0 to \(8\pi\), as this represents one period of the function. Mark the midpoint of the period at \(4\pi\), and the quarter points at \(2\pi\) and \(6\pi\). The y-values will range from -2 to 2 (negative to positive amplitude). The graph starts at y=0 at x=0 (origin), reaches the maximum amplitude at \(2\pi\), returns to 0 at \(4\pi\), goes to the minimum amplitude at \(6\pi\), and finally back to 0 at \(8\pi\), completing one period.