Problem 44 - Precalculus Essentials

Question

Graph two periods of the given cosecant or secant function. $$y=2 \sec \left(x+\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The graph of the function \(y=2 \sec \left(x+\frac{\pi}{2}\right)\) includes vertical asymptotes at \(x = \frac{\pi}{2} + n\pi\) and max/min points at \(x= n\pi\), where \(n\) is an integer. The curve connected these points, creating two complete periods of the function.

Step 1: Understand the Function

The given function is \(y=2 \sec \left(x + \frac{\pi}{2}\right)\). The \(2\) in front of the secant function indicates a vertical stretch by a factor of 2. The \(\frac{\pi}{2}\) added to \(x\) indicates a horizontal shift to the left by \(\frac{\pi}{2}\) units.

Step 2: Identify the Key Features of the Graph

Because we’re working with a secant function, the key features to note are the vertical asymptotes and the maximum and minimum points. The standard secant function has vertical asymptotes at \(x= n\pi\), where \(n\) is an odd integer, and has maximum and minimum points midway between the vertical asymptotes. Noting the transformation in our function, the vertical asymptotes are shifted to \(\frac{\pi}{2} + n\pi\) and the max/min points are at \(x= n\pi\), where \(n\) is an integer.

Step 3: Plot the Key Features on the Graph

Place vertical asymptote lines at \(x = \frac{\pi}{2} + n\pi\), where \(n\) is an integer. Mark the maximum and minimum points at \(x= n\pi\), where \(n\) is an integer. The maximum and minimum values will vary depending on whether \(n\) is odd or even due to the wave nature of the secant function.

Step 4: Draw the Graph

Connect the maximum and minimum points with a smooth curve moving towards the vertical asymptotes but never crossing them. Continue this pattern until two complete periods of the function are graphed.