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Chapter 4: Problem 57

Use a vertical shift to graph one period of the function. $$y=2 \sin \frac{1}{2} x+1$$

Short Answer

Expert verified
The graph of the function \(y=2 \sin\ (\frac{1}{2} x)+1\) has an amplitude of 2, frequency of \(\frac{1}{2}\), period of \(4\pi\) and a vertical shift of 1. The graph starts and ends at y=1 and reaches a peak at y=3 and a trough at y=1.

Step by step solution

01

Identify the Amplitude, Frequency, Period and Vertical Shift

The given function is of the form \(y=A \sin(Bx) + D\). Here, the amplitude 'A' is the absolute value of the coefficient of the sine function, which is 2. The frequency 'B' is the coefficient of x inside the sine function, which is \(\frac{1}{2}\). The period is equal to \( \frac{2 \pi}{|B|} = 4\pi \). The vertical shift 'D' is the term added or subtracted at the end, which is 1 in this case.
02

Determine the Start and End of the Period

Since there is no phase shift, the period starts at x=0. Since the period of this function is \(4\pi\), it ends at x=\(4\pi\).
03

Construct the Graph

On the x-axis, mark the start and end of the period (0 to \(4\pi\)). Next, note down the vertical shift which moves the graph up by 1 unit. Now, plot the five key points - start, peak, mid, trough, end. The start and end will be on the midline (y=1), the peak will be 1 unit above the midline (y=3), and the trough will be 1 unit below the midline (y=1). Draw a smooth curve through these points to represent one period of the function.

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