Problem 43 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The graph of \(y=2 \sec (x+\pi)\) is the graph of the secant function with an amplitude of 2 and a phase shift of \(\pi\), with asymptotes at multiples of \(\pi\). The graph is sketched for two periods, between \(-2\pi\) and \(2\pi\).

Step 1: Identify amplitude and phase shift

The amplitude of this function is 2 because \(|2|\) (the absolute value of 2) is 2. The phase shift is \(\pi\) because that is the value being added to \(x\) inside the parentheses of the secant function. This means that the graph is shifted \(\pi\) units to the left.

Step 2: Plot related function

Before graphing the secant function, it's simpler to plot the related cosine function: \(y=2 \cos (x+\pi)\). The cos function oscillates between -2 and 2, with period \(2\pi\). In order to find two periods plot function on the interval \(-2\pi \leq x \leq 2\pi\). The graph of this function will be an oscillation that starts at a minimum (since it is shifted \(\pi\) units to the left), and goes up to the maximum, then comes back down.

Step 3: Plot secant function

After plotting the cosine function, the next thing to do is to place the asymptotes at the points where the cosine function crosses the axis, since secant is undefined where cosine equals zero. Then at the peaks and troughs of the related cosine graph make points of secant graph. Between these points draw \(U\) shaped curves forming the secant graph.

Step 4: Draw two periods of secant

After plotting the secant function for one period, which is on the interval \(-\pi \leq x \leq \pi\), repeat the same process for another period on the interval \(\pi \leq x \leq 3\pi\) to plot two complete periods of the secant function. Label the extreme points and asymptotes on the graph.