Sure! Let's find the amplitude and period of the function [tex]\( y = 4 \sin \left(\frac{2}{3} x \right) \)[/tex].
Firstly, to determine the amplitude, we look at the coefficient in front of the sine function. The given function is of the form [tex]\( y = A \sin(Bx) \)[/tex].
In this case, the coefficient [tex]\( A \)[/tex] is 4. Therefore, the amplitude is:
[tex]\[
\text{Amplitude} = 4
\][/tex]
Next, to determine the period of the sine function, we use the formula for the period of [tex]\( y = \sin(Bx) \)[/tex], which is given by:
[tex]\[
\text{Period} = \frac{2\pi}{B}
\][/tex]
Here, [tex]\( B \)[/tex] is the coefficient of [tex]\( x \)[/tex] inside the sine function. In our function, [tex]\( B = \frac{2}{3} \)[/tex].
Substituting [tex]\( B \)[/tex] into the period formula, we get:
[tex]\[
\text{Period} = \frac{2\pi}{\frac{2}{3}} = 2\pi \times \frac{3}{2} = 3\pi
\][/tex]
Thus, the period of the function is:
[tex]\[
\text{Period} = 3\pi
\][/tex]
So, we have the final solutions:
[tex]\[
\begin{aligned}
\text{Amplitude} &= 4 \\
\text{Period} &= 3\pi
\end{aligned}
\][/tex]
Therefore, the exact values for the period and amplitude of the function [tex]\( y = 4 \sin \left( \frac{2}{3} x \right) \)[/tex] are:
Period: [tex]\( 3\pi \)[/tex]
Amplitude: [tex]\( 4 \)[/tex]
