Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 645: 73

Answer

The amplitude of the function is $3$ and period is $6\pi $.

Work Step by Step

The given equation is in the form of $y=A\cos Bx$. Here, $A=3\text{ and }B=\frac{1}{3}$ So, the amplitude is: $\begin{align} & \text{Amplitude=}\left| A \right| \\ & =\left| 3 \right| \\ & =3 \end{align}$ The period is given below: $\begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{\left( \frac{1}{3} \right)} \\ & =6\pi \end{align}$ $\text{Phase shift=}\frac{C}{B}=0$ And the quarter period is as follows: $\begin{align} & \text{Quarter-period}=\frac{6\pi }{4} \\ & =\frac{3\pi }{2} \end{align}$ Now, add quarter periods starting from $x=\text{0}$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=\text{0}$ And the x-value for the second key point is: $\begin{align} & x=0+\frac{3\pi }{2} \\ & =\frac{3\pi }{2} \end{align}$ And the x-value for the third key point is: $\begin{align} & x=\frac{3\pi }{2}+\frac{3\pi }{2} \\ & =3\pi \end{align}$ And the x-value for the fourth key point is: $\begin{align} & x=3\pi +\frac{3\pi }{2} \\ & =\frac{9\pi }{2} \end{align}$ And the x-value for the fifth key point is: $\begin{align} & x=\frac{9\pi }{2}+\frac{3\pi }{2} \\ & =6\pi \end{align}$
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