Problem 57 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The graph of the function \(y=2 \sin\left(\frac{1}{2}x\right) + 1\) will be a sine wave with an amplitude of 2, a period of \(4\pi\) radians, and shifted up by 1 unit from the usual sine function.

Step 1: Determine the Period

The period \(P\) of a sine function of the form \(y = a \sin(bx)\) is given by \(P = \frac{2\pi}{|b|}\). For the function \(y = 2 \sin\left(\frac{1}{2}x\right) + 1\), \(b = \frac{1}{2}\), so the period is \(P = \frac{2\pi}{\frac{1}{2}} = 4\pi\) radians.

Step 2: Graph the Sine Function with the Amplitude

The original sine function \(y = \sin x\) oscillates between -1 and 1. A coefficient \(a\) alters the amplitude, or the distance from the maximum and minimum to the average value (normally, the x-axis). In this case, \(a = 2\), so the maximum will be at 2 and the minimum at -2. Draw the basic sine curve over one period (from 0 to \(4\pi\)), making sure the maximum occurs at 2 and the minimum at -2.

Step 3: Apply the Vertical Shift

The '+1' outside the sine function indicates a vertical shift upwards by 1 unit. This means each point on the sine curve drawn in Step 2 should be shifted up by 1 unit. The maximum will now be at 3 and the minimum at -1.