## Precalculus (6th Edition) Blitzer

Equation (1) is of the form $y=A\sin \left( Bx-C \right)$. We can undertake the following steps to determine the graph of the equation: Step 1. Here, $A=3,B=2$, and $C=\pi$, such that Amplitude $=3$ \begin{align} & \text{Period}=\frac{2\pi }{B} \\ & =\frac{2\pi }{2} \\ & =\pi \end{align} \begin{align} & \text{Phase shift}=\frac{C}{B} \\ & =\frac{\pi }{2} \end{align} $\text{Quarter period}=\frac{\pi }{4}$ Step 2. Now we will find the $x$ -values by adding the quarter periods. \begin{align} & x=\frac{\pi }{2} \\ & x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4} \\ & x=\frac{3\pi }{4}+\frac{\pi }{4}=\pi \\ & x=\pi +\frac{\pi }{4}=\frac{5\pi }{4} \end{align} Step 3. We will evaluate the function at each value of $x$. At $x=\frac{\pi }{2}$, \begin{align} & y=3\sin \left( 2\cdot \frac{\pi }{2}-\pi \right) \\ & =3\sin \left( 0{}^\circ \right) \\ & =0 \end{align} Therefore, the coordinates are $\left( \frac{\pi }{2},0 \right)$. At $x=\frac{3\pi }{4}$, \begin{align} & y=3\sin \left( 2\cdot \frac{3\pi }{4}-\pi \right) \\ & =3\sin \left( \frac{\pi }{2} \right) \\ & =3 \end{align} Therefore, the coordinates are $\left( \frac{3\pi }{4},3 \right)$. At $x=\pi$, \begin{align} & y=3\sin \left( 2\cdot \pi -\pi \right) \\ & =3\sin \left( \pi \right) \\ & =0 \end{align} Therefore, the coordinates are $\left( \pi,0 \right)$. At $x=\frac{5\pi }{4}$, \begin{align} & y=3\sin \left( 2\cdot \frac{3\pi }{2}-\pi \right) \\ & =3\sin \left( 2\pi \right) \\ & =0 \end{align} Therefore, the coordinates are $\left( \frac{3\pi }{2},0 \right)$. Step 4. On connecting these points, the graph can be obtained. Step 5. Using all the steps above, we produce the final graph.