Problem 47 - Precalculus Essentials

Question

Graph two periods of each function. $$y=\sec \left(2 x+\frac{\pi}{2}\right)-1$$

Step-by-Step Solution

Answer

The graph of the function \(y=\sec \left(2 x+\frac{\pi}{2}\right)-1\) has a period of \(\pi\), it is shifted to the left by \(\frac{\pi}{4}\), and it is shifted downward by \(1\) unit. It should be plotted between \(-\frac{\pi}{4}\) and \( \frac{7\pi}{4}\) for graphing two periods.

Step 1: Identify the period

The period of a secant function is determined by the coefficient of \(x\) inside the secant function. In our equation, the coefficient of \(x\) is \(2\). The period of the base \(sec(x)\) function is \(2\pi\), therefore, the period of \(sec(2x)\) is \(\pi\). So, one complete cycle of this function will occur over the interval \(0\) to \(\pi\). To get two periods, we must graph from \(0\) to \(2\pi\).

Step 2: Identify the phase shift (or horizontal shift)

The phase shift is determined by the term added or subtracted from \(x\) inside the secant function. In our equation, \( \frac{\pi}{2}\) is added to \(2x\), shifting the graph to the left by \(\frac{\pi}{4}\) (divide by the coefficient of \(x\) to obtain phase shift). Thus, our graphing interval will be \(-\frac{\pi}{4}\) to \( \frac{7\pi}{4}\) (shifting \(0\) to \(2\pi\) to the left by \(\frac{\pi}{4}\))

Step 3: Identify the vertical shift

The vertical shift is determined by the term added or subtracted from the secant function. In our equation, \(1\) is subtracted, shifting the graph down by \(1\) unit. Therefore, the secant vertical asymptote from the base function will shift down by \(1\).

Step 4: Plot the function

Now, refer to the standard secant graph (or graph of \(y=cos(x)\) and then perform inverse function for reference) and plot the function from \(-\frac{\pi}{4}\) to \( \frac{7\pi}{4}\) considering the vertical shift down by \(1\) unit and vertical asymptotes shifted correspondingly.