Problem 32 - Precalculus Essentials

Question

Graph two periods of the given cosecant or secant function. $$y=\frac{3}{2} \csc \frac{x}{4}$$

Step-by-Step Solution

Answer

The graph of the function \(y=\frac{3}{2} \csc \left(\frac{x}{4}\right)\) has amplitude of 3/2, period of \(8\pi\), and asymptotes at \(x = 0, \pm 4n\pi\) (for all integer values of n).

Step 1: Identity the Function’s Form

The function \(y=\frac{3}{2} \csc \left(\frac{x}{4}\right)\) is in the form of \(y=A \csc(Bx)\), where A is the amplitude and B affects the period. In this function, A = 3/2 and B = 1/4.

Step 2: Calculate Amplitude

The amplitude of the function is |A|. In our case, it is \(\left|\frac{3}{2}\right| = \frac{3}{2}\). Amplitude is a measure of how high and low the graph goes from its midline.

Step 3: Calculate Period

The period of a csc function is \(T=2\pi|\frac{1}{B}|\). Here, B = 1/4, so the period is \(T = 2\pi|\frac{1}{1/4}|= 2\pi*4 = 8\pi\). This is the width of one complete cycle on the graph.

Step 4: Identify the Asymptotes

The function \(y=csc(x)\) is undefined wherever \(sin(x) = 0\). Thus, the function \(y=\frac{3}{2} \csc \left(\frac{x}{4}\right)\) will have a vertical asymptote wherever \(sin(\frac{x}{4}) = 0\), which occurs at \(x = 0, \pm 4n\pi\), for all integer values of n.

Step 5: Draw the Graph

Start at \(x=0\) (where the first asymptote is). Mark the period intervals along the x-axis (each \(8\pi\)). Draw an upward curve reaching its max at the central point between two asymptotes where the y value is 3/2 and downward curve where the y value is -3/2. Repeat these steps for two periods. This way, the graph of the function is drawn.