Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.3 - Trigonometric Graphs - 5.3 Exercises - Page 429: 49

Answer

a. amplitude=$\displaystyle \frac{3}{2}$, period=$\displaystyle \frac{2\pi}{3}$, horizontal shift=0 b. $y=\displaystyle \frac{3}{2}\cos 3x$

Work Step by Step

For $y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$, $(k>0)$, Amplitude: $|a|$, Period: $\displaystyle \frac{2\pi}{k}$, Horizontal shift: $b$ ----------- f(0)$\neq$0, its maximum is at x=0, and after x=0, the graph descends. These are characteristics of a cosine curve with positive a, and no horizontal shift (b=0). The amplitude is $\displaystyle \frac{3}{2}$, so a can be $\displaystyle \pm\frac{3}{2}$, but since it is positive, a=$\displaystyle \frac{3}{2}$. The period is $\displaystyle \frac{4\pi}{6}=\frac{2\pi}{3}$ (the next unmarked value on the x-axis after $\displaystyle \frac{3\pi}{6}=\frac{\pi}{2}$) So, from $\displaystyle \frac{2\pi}{k} =\frac{2\pi}{3}$, it follows that k=$3$. With these parameters, $f(x)=y=a\cos k(x-b)$ $y=\displaystyle \frac{3}{2}\cos 3x$ a. amplitude=$\displaystyle \frac{3}{2}$, period=$\displaystyle \frac{2\pi}{3}$, horizontal shift=0 b. $y=\displaystyle \frac{3}{2}\cos 3x$
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