Chapter 6.6, Problem 3AYU

To determine

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

Answer to Problem 3AYU

Solution:

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

Explanation of Solution

Given:

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

Calculation:

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

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

One cycle will begin at $x=\frac{\varphi }{\omega }=\frac{\pi }{2}$ and end at $x=\frac{\varphi }{\omega }+\frac{2\pi }{\omega }=\frac{\pi }{2}+\pi =\frac{3\pi }{2}$.

To find five key points, divide the interval $\left[\frac{\pi }{2}, \frac{3\pi }{2}\right]$ in to four sub intervals, each of length $\frac{\frac{3\pi }{2}-\frac{\pi }{2}}{4}=\frac{\pi }{4}$.

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

$\left(\frac{\pi }{2}, 0\right), \left(\frac{3\pi }{4}, 4\right), \left(\pi , 0\right), \left(\frac{5\pi }{4}, -4\right), \left(\frac{3\pi }{2}, 0\right)$

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