Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Review Exercises - Page 190: 37

Answer

Refer to the graph below.
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Work Step by Step

RECALL: (1) The function $y=a\cdot \sec{x}$ has a period of $2\pi$. (2) Consecutive asymptotes of the secant function are $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ (3) The function $y=a \cdot \sec{[b(x-d)]}$ has a period of $\frac{2\pi}{|b|}$ and involves a phase shift of $|d|$ (to the right when $d\gt0$, to the left when $d\lt0$). Write the given equation in the form $y=a \cdot \sec{[b(x-d)]}$ by factoring out $2$ within the secant function to obtain: $y=\sec{[2(x+\frac{\pi}{6})]}$ Thus, the given function has $a=1, b=2, $and $d=-\frac{\pi}{6}$. This means that it has: period = $\frac{2\pi}{2}=\pi$ phase shift = $|-\frac{\pi}{6}|=\frac{\pi}{6}$, to the left The guide function for this secant is $y=\cos{[2(x+\frac{\pi}{6})]}$. One period of the function $y=\sec{(2x)}$ is in the interval $[0, 2\pi]$. This means that one period of the function $y=\sec{[2(x+\frac{\pi}{6})]}$, which involves a $\frac{\pi}{6}$ shift to the left, will be at $[-\frac{\pi}{6}, \frac{5\pi}{6}]$. Divide this interval into four equal parts to get the key x-values $-\frac{\pi}{6}, \frac{\pi}{12}, \frac{\pi}{3}, \frac{7\pi}{12}, \frac{5\pi}{6}$. Find the consecutive vertical asymptotes by equating $2x+\frac{\pi}{3}$ to $\frac{\pi}{2}$ and to $\frac{3\pi}{2}$, then solve each equation to obtain: \begin{array}{ccc} 2x+\frac{\pi}{3}&=\frac{\pi}{2} &\text{or} &2x+\frac{\pi}{3} = \frac{3\pi}{2} \\2x&=\frac{\pi}{6} &\text{or} &2x=\frac{7\pi}{6} \\x&=\frac{\pi}{12} &\text{or} &x=\frac{7\pi}{12} \end{array} To graph the given function, perform the following steps: (1) Create a table of values for the guide function $y=\cos{[2(x+\frac{\pi}{6}]}$ using the key x-values listed above. (Refer to the table below.) (2) Plot the points from the table of values and connect them using a dashed curve (as the curve will only serve as a guide). (3) Graph the consecutive vertical asymptotes $x=\frac{\pi}{12}$ and $x=\frac{7\pi}{12}$. (4) Sketch the graph of $y=\sec{[2(x+\frac{\pi}{6})]}$ by drawing (i) a U-shaped curve below the x-axis and between the consecutive vertical asymptotes. (ii) a half U-shaped curve from the point $(-\frac{\pi}{6}, 1)$ to the asymptote $x=\frac{\pi}{12}$. (iii) a half U-shaped curve from the asymptote $x=\frac{7\pi}{12}$ to the point $(\frac{5\pi}{6}, 1)$ . (Refer to the graph in the answer part above.)
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