Chapter 6.6, Problem 12AYU

In Problems 3-14, find the amplitude, period, and phase shift of each function. Graph each function. Be sure to label key points. Show at least two periods.

$y\text{ = }2\text{cos}\left(2\pi x-4\right)-1$

To determine

To find: The amplitude, period, phase shift of the function, graph the function.

Answer to Problem 12AYU

Solution:

The amplitude $=2$, period $=1$, phase shift $=\frac{2}{\pi }$.

Explanation of Solution

Given:

$y=2\cos \left(2\pi x-4\right)-1$

Calculation:

Compare $y=2\cos \left(2\pi x-4\right)-1$ to $y=A\cos \left(\omega x-\varphi \right)+B$, note that $A=2, \omega =2\pi$, $\varphi =4$ and $B=-1$. The graph is a cosine curve with amplitude $\left|A\right|=2$, period $T=\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=1$ and phase shift $=\frac{\varphi }{\omega }=\frac{4}{2\pi }=\frac{2}{\pi }$.

The graph of $y=2\cos \left(2\pi x-4\right)-1$ will lie between $-2$ and 2 on the $y\text{-axis}$.

One cycle will begin at $x=\frac{\varphi }{\omega }=\frac{2}{\pi }$ and end at $x=\frac{\varphi }{\omega }+\frac{2\pi }{\omega }=\frac{2}{\pi }+1=\frac{2+\pi }{\pi }$.

To find five key points, divide the interval $\left[\frac{2}{\pi },\frac{2+\pi }{\pi }\right]$ in to four sub intervals, each of length $\frac{\frac{2+\pi }{\pi }-\frac{2}{\pi }}{4}=\frac{1}{4}$.

Use the values of $x$ to determine the five key points on the graph:

$x$	$y=2\cos \left(2\pi x-4\right)$
$\frac{2}{\pi }$	$2$
$\frac{8+\pi }{4\pi }$	$0$
$\frac{8+2\pi }{4\pi }$	$-2$
$\frac{8+3\pi }{4\pi }$	$0$
$\frac{2+\pi }{4\pi }$	$2$

$x$	$y=2\cos \left(2\pi x-4\right)-1$
$\frac{2}{\pi }$	$1$
$\frac{8+\pi }{4\pi }$	$-1$
$\frac{8+2\pi }{4\pi }$	$-3$
$\frac{8+3\pi }{4\pi }$	$-1$
$\frac{2+\pi }{\pi }$	$1$

Plot these five points and fill in the graph of the sine function.

$y=2\cos \left(2\pi x-4\right)$	$y=2\cos \left(2\pi x-4\right)-1$