## Precalculus (6th Edition) Blitzer

Amplitude of the given function is $3$, period is $\pi$, and phase shift is $\frac{\pi }{2}$.
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 $y=3\cos \left( 2x-\pi \right)$