Answer
Amplitude of the given function is $3$, period is $\pi $, and phase shift is $\frac{\pi }{2}$.
Work Step by Step
For the standard function in the form $y=A\cos \left( Bx+C \right)$
$\begin{align}
& \text{Amplitude}=\left| A \right| \\
& \text{Period}=\frac{2\pi }{\left| B \right|}
\end{align}$
And,
$\text{Phase}\,\text{Shift}=-\frac{C}{B}$
Now, compare the given function $y=3\cos \left( 2x-\pi \right)$ with the standard function as follows:
$A=3,\text{ }B=2,\text{ and }C=-\pi $
So,
$\begin{align}
& \text{Amplitude}=\left| 3 \right| \\
& =3 \\
& \text{Period}=\frac{2\pi }{\left| 2 \right|} \\
& =\pi
\end{align}$
And,
$\begin{align}
& \text{Phase}\,\text{Shift}=-\frac{\left( -\pi \right)}{2} \\
& =\frac{\pi }{2}
\end{align}$
Then, calculate the quarter period as follows:
$\begin{align}
& \text{Quarter}\,\text{Period}=\frac{\text{Period}}{4} \\
& =\frac{\pi }{4}
\end{align}$
Thus, the cycle begins at $x=\frac{\pi }{2}$. Find the x-value and y-value at multiples of the quarter period.
Now, plot the obtained coordinates to get the graph of the function $y=3\cos \left( 2x-\pi \right)$
