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Chapter 4: Problem 35

Sketch the graph of the function. (Include two full periods.) $$y=2 \sec (x+\pi)$$

Short Answer

Expert verified
The graph of the function \(y=2 \sec (x + \pi)\) is a series of U and inverted U shapes, centered along the x-axis, with opening widths of \(2\pi\), and with vertical asymptotes at points of the form \(x = (2n+1)\pi/2 - \pi\).

Step by step solution

01

Define the Base Function

Secant is the reciprocal of cosine, namely \(\sec(x) = 1/\cos(x)\). The cosine function oscillates between -1 and 1. Its graph is a continuous, smooth wave. However, secant has vertical asymptotes at those values of x where cosine is zero.
02

Determine Asymptotes

The cosine function is zero at odd multiples of \(\pi/2\). Therefore, the secant function has vertical asymptotes at these points. For the function \(y=2 \sec (x + \pi)\), this means that asymptotes occur at x-values satisfying \(x + \pi = (2n+1)\pi/2\), where n is an integer, i.e., \(x = (2n+1)\pi/2 - \pi\). This gives us the general location of the vertical asymptotes.
03

Identify Key Points

When \(x + \pi = n\pi\), where n is an integer, the cosine function equals \(-1\) or \(1\), and thus secant equals -1 or 1. This gives the points on the graph of the secant function closest to the vertical asymptotes. For our function \(y=2 \sec (x + \pi)\), these points occur when \(x = n\pi - \pi\), with y-coordinates 2 or -2.
04

Draw the Graph

Now with these points and asymptotes, sketch the graph of the function. It repeats every \(2\pi\) units, so to include two full periods, simply repeat the same shape over an interval of \(4\pi\).

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