Chapter 6.6, Problem 9AYU

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.

$y\text{ = }4\text{sin}\left(\pi x+2\right)-5$

To determine

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

Answer to Problem 9AYU

Solution:

The amplitude $=4$, period $=4$, phase shift $=-\frac{2}{\pi }$.

Explanation of Solution

Given:

$y=4\sin \left(\pi x+2\right)-5$

Calculation:

Compare $y=4\sin \left(\pi x+2\right)-5$ to $y=A\sin \left(\omega x-\varphi \right)+B$, note that $A=4, \omega =\pi$, $\varphi =-2$ and $B=-5$. The graph is a sine curve with amplitude $\left|A\right|=4$, period $T=\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2$ and phase shift $=\frac{\varphi }{\omega }=\frac{-2}{\pi }=-\frac{2}{\pi }$.

The graph of $y=4\sin \left(\pi x+2\right)-5$ will lie between $-4$ and 4 on the $y\text{-axis}$.

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

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

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

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

$x$	$y=4\sin \left(\pi x+2\right)-5$
$-\frac{2}{\pi }$	$-5$
$\frac{-4+\pi }{2\pi }$	$-1$
$\frac{-4+2\pi }{2\pi }$	$-5$
$\frac{-4+3\pi }{2\pi }$	$-9$
$\frac{-2+2\pi }{\pi }$	$-5$

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

$y=4\sin \left(\pi x+2\right)$	$y=4\sin \left(\pi x+2\right)-5$