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Chapter 4: Problem 35

Graph two periods of the given cosecant or secant function. $$y=\sec \frac{x}{3}$$

Short Answer

Expert verified
The graph of \(y=\sec \frac{x}{3}\) is a series of U-shaped curves, each approaching vertical asymptotes at \(x=-5\pi\), \(x=-\pi\), \(x=\pi\), \(x=5\pi\), \(x=11\pi\), and \(x=17\pi\) for two periods. The curves pass through the maxima and minima of the cosine graph which are its x-intercepts. These are located halfway between the vertical asymptotes. The y-intercept is at the point \((0, 1)\)

Step by step solution

01

Define the Cosine function

Since the function given is \(y=\sec \frac{x}{3}\), we can define the corresponding cosine function as \(y=\cos \frac{x}{3}\). This will allow us to find the points at which the cosine function crosses the x-axis and hence determine the vertical asymptotes for the graph of the secant function.
02

Identify the period of the function

Next, calculate the period of the function. The period of the standard secant function \(y = sec(x)\) is \(2\pi\). In this case, the input to the secant function is divided by 3, such that there are 3 cycles in where there was one before. Hence, the period of \(y = sec(\frac{x}{3})\) is \(6\pi\).
03

Draw the graph of the cosine function

Graph the cosine function, \(y=\cos \frac{x}{3}\), for two periods, taking account of its period, which is \(6\pi\).
04

Identify the vertical asymptotes

To find the vertical asymptotes of the secant graph, find the x-values for which the cosine function is equal to zero. There are vertical asymptotes at these x-values. The vertical asymptotes for one period of the secant function are at \(x=\pi\) and \(x=5\pi\). Therefore, the vertical asymptotes for two periods will also occur at \(x=-5\pi\), \(x=-\pi\), \(x=\pi\), \(x=5\pi\), \(x=11\pi\), and \(x=17\pi\). Any secant graph is undefined at these points.
05

Draw the secant function

The secant graph will approach asymptotes, passing through maxima and minima of the cosine graph which are its x-intercepts. The x-intercepts are consistently located halfway between the vertical asymptotes. The y-intercept is at \((0, 1)\).

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Most popular questions from this chapter

Use the keys on your calculator or graphing utility for converting an angle in degrees, minutes, and seconds \(\left(D^{\circ} M^{\prime} S^{\prime \prime}\right)\) into decimal form, and vice versa. Convert each angle to a decimal in degrees. Round your answer to two decimal places. $$30^{\circ} 15^{\prime} 10^{\prime \prime}$$

Graph \(f, g,\) and \(h\) in the same rectangular coordinate system for \(0 \leq x \leq 2 \pi .\) Obtain the graph of h by adding or subtracting the corresponding \(y\) -coordinates on the graphs of \(f\) and \(g\) $$f(x)=-2 \sin x, g(x)=\sin 2 x, h(x)=(f+g)(x)$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. When graphing one complete cycle of \(y=A \cos (B x-C)\) I find it easiest to begin my graph on the \(x\) -axis.

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=-3 \cos \left(2 x-\frac{\pi}{2}\right)$$

Describe a general procedure for obtaining the graph of \(y=A \sin (B x-C)\)
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