Trigonometry (11th Edition) Clone

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

Chapter 4 - Graphs of the Circular Functions - Section 4.3 Graphs of the Tangent and Cotangent Functions - 4.3 Exercises - Page 171: 27

Answer

Refer to the graph below.
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Work Step by Step

RECALL: (1) The function $y=\cot{[b(x-d)]}$ has: period = $\frac{\pi}{b}$ phase (horizontal) shift = $|d|$, (to the right when $d\gt 0$, to the left when $d\lt0$) (2) The consecutive vertical asymptotes of the function $y=\cot{x}$, whose period is $\pi$, are $x=0$ and $x=\pi$. Write the given funtion in $y=\cot{[b(x-d)]}$ form by factoring out $3$ within the cotangent function to obtain: $y=\cot{[3(x+\frac{\pi}{12})]}$ Thus, with $b=3$ and $d=-\frac{\pi}{12}$, the given function has: period = $\frac{\pi}{3}$ phase (horizontal) shift = $\frac{\pi}{12}$ to the left To find the consecutive vertical asymptotes, equate $(3x+\frac{\pi}{4})$ to $0$ and to $\pi$, then solve each equation to obtain: \begin{array}{ccc} 3x+\frac{\pi}{4}&=0 &\text{ or } &3x+\frac{\pi}{4}=\pi \\3x&=-\frac{\pi}{4} &\text{or} &3x=\frac{3\pi}{4} \\x&=-\frac{\pi}{12} &\text{or} &x=\frac{\pi}{4} \end{array} This means that one period of the given function is in the interval $[-\frac{\pi}{12}, \frac{\pi}{4}]$ and the next period is $[\frac{\pi}{4}, \frac{7\pi}{12}]$. Dividing each interval into four equal parts give the key x-values: $0, \frac{\pi}{12}, \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{\pi}{2}$. To graph the given function, perform the following steps: (1) Create a table of values for the given function using the key x-values listed above. (Refer to the attached image table below.) (2) Graph the consecutive vertical asymptotes. (3) Plot each point from the table then connect them using a smooth curve, making sure that the curves are asymptotic with the lines in Step (2) above. Refer to the graph in the answer part above.
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