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Chapter 4: Problem 37

Sketch one full period of the graph of each function. $$y=-2 \sec \pi x$$

Short Answer

Expert verified
The graph of \(y=-2 \sec \pi x\) is a reflection of the secant function over the x-axis and has period 2. It has vertical asymptotes at \(x= \pm \frac{1}{2}, \pm \frac{3}{2}\), etc., and x-intercepts at \(x=0\) and \(x=1\).

Step by step solution

01

Identify the Period

The period of the secant function is normally \(2\pi\) but since there is a coefficient of \(\pi\) in the argument of the function, the period of \(y=-2 \sec \pi x\) would be \(\frac{2\pi}{|\pi|} = 2\).
02

Identify the Asymptotes

The cosine function has zeros at \(\pm \frac{1}{2}, \pm \frac{3}{2}\), etc., within the period from \(0\) to \(2\), so the secant function will have vertical asymptotes at these points within each period.
03

Identify the x-Intercepts

The cosine function has a maximum of \(1\) at \(0\) and a minimum of \(-1\) at \(1\) within each period. So, the secant function will have x-intercepts at \(x=0\) and \(x=1\).
04

Sketch the Graph

Field the points and asymptotes on the graph. The graph of \(-2\sec \pi x\) is a reflection of the secant function over the x-axis. Draw smooth curves connecting the points, approaching but never crossing the asymptotes.

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