Problem 40 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

To graph two periods of the function \(y=-\frac{3}{2} \sec \pi x\), one can graph the related function \(y=-\frac{3}{2} \cos \pi x\) first. Then, add vertical asymptotes where the cosine function equals zero, and sketch the secant function based on those asymptotes and the graph of the cosine function.

Step 1: Understand the basic secant function graph

A basic \(sec(x)\) function has undefined values (unreachable) at every \(x\) value where \(cos(x) = 0\). Therefore, it has vertical asymptotes at \(x = \frac{(2n+1)\pi}{2}\), where \(n\) is an integer. They usually have a period of \(2\pi\) and the range is \(y \leq -1\) or \(y \geq 1\).

Step 2: Draw the related cosine function

The function \(y=-\frac{3}{2} \sec \pi x\) is equivalent to \(y=-\frac{3}{2} \cos \pi x\). The coefficient of \(x\), \(\pi\), indicates a horizontal compression of the graph by a factor of \(\frac{1}{\pi}\), or a change in the period from \(2\pi\) to \(2\). The negative sign gives a reflection of the graph across the x-axis. The factor of \(-\frac{3}{2}\) means a vertical stretch of the base graph by a factor of \(\frac{3}{2}\). Draw this function for 2 periods first.

Step 3: Draw the secant graph based on the cosine graph

Now you can draw the graph of the given secant function based on the cosine graph. At points where the cosine function equals to zero, draw vertical asymptotes for the secant function. The secant graph will diverge at these points. For the points in between, the secant graph will leave from the cosine graph, reach a maximum or minimum, then approach the cosine graph again. Repeat this process over two periods of the function.