Chapter 6, Problem 74RE

To determine

Determine the amplitude, period and phase shift of each function and graph at least two periods.

Answer to Problem 74RE

Amplitude of $y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)$ is $\boxed{amplitude=7}$

Periodicity of $y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)$ is

  $\boxed{periodicity=6}$

Phase shift of $y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)$ is $\boxed{\frac{4}{\pi }}$

Explanation of Solution

Given information:

  $y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)$

Formula used:

For $f\left(x\right)=a.\sin \left(bx-c\right)+d$ the amplitude is $\left|a\right|$

Periodicity of $a.\sin \left(bx-c\right)+d=\frac{\text{Periodicity of }\sin x}{\left|b\right|}$

For $f\left(x\right)=A\cdot g\left(Bx-C\right)+D$ , where $g\left(x\right)$ is one the basic trigonometric functions, $\frac{C}{B}$ is phase shift and D is vertical shift.

Now calculating the amplitude of the function by using $f\left(x\right)=a.\sin \left(bx-c\right)+d$ the amplitude is $\left|a\right|$ formula:

  $\begin{array}{l}y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)\\ Here,a=-7\\ \\ amplitude=\left|a\right|\\ amplitude=\left|-7\right|\\ \boxed{amplitude=7}\end{array}$

Now calculating periodicity by using $a\cdot \sin \left(bx+c\right)+d=\frac{\text{Periodicity of }\sin x}{\left|b\right|}$ formula:

  $\frac{\text{Periodicity of sinx}}{\left|b\right|}$

Periodicity of $\sin x$ is $2\pi$ and $b=\frac{\pi }{3}$ , therefore:

  $\begin{array}{l}\\ =\frac{2\pi }{\left|\frac{\pi }{3}\right|}\\ \\ \boxed{periodicity=6}\end{array}$

Now calculating the phase shift of $y=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)$ :

  $\begin{array}{l}f\left(x\right)=-7\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right)\\ \\ g\left(Bx-C\right)=\sin \left(\frac{\pi }{3}x-\frac{4}{3}\right),B=\frac{\pi }{3},C=\frac{4}{3},D=0\end{array}$

Therefore, the phasing shift $\frac{C}{B}$ is $\boxed{\frac{\frac{4}{3}}{\frac{\pi }{3}}=\frac{4}{\pi }}$