Question

graph one period of y = 3sin2x

Explanation:

Step1: Find the period

For the function $y = A\sin(Bx)$, the period $T=\frac{2\pi}{B}$. Here $B = 2$, so $T=\frac{2\pi}{2}=\pi$.

Step2: Determine key - points

The standard key - points for $y=\sin x$ in one period $[0, 2\pi]$ are $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2}, - 1),(2\pi,0)$. For $y = 3\sin(2x)$, we find the $x$ - values by setting $2x$ equal to the standard $x$ - values of $\sin x$ and solving for $x$.
When $2x = 0$, $x = 0$ and $y=3\sin(0)=0$.
When $2x=\frac{\pi}{2}$, $x=\frac{\pi}{4}$ and $y = 3\sin(\frac{\pi}{2})=3$.
When $2x=\pi$, $x=\frac{\pi}{2}$ and $y=3\sin(\pi)=0$.
When $2x=\frac{3\pi}{2}$, $x=\frac{3\pi}{4}$ and $y=3\sin(\frac{3\pi}{2})=-3$.
When $2x = \pi$, $x=\frac{\pi}{2}$ and $y = 3\sin(\pi)=0$.

Step3: Plot the points

Plot the points $(0,0),(\frac{\pi}{4},3),(\frac{\pi}{2},0),(\frac{3\pi}{4}, - 3),(\pi,0)$ and connect them with a smooth curve to graph one period of $y = 3\sin(2x)$.

Answer:

Graph the points $(0,0),(\frac{\pi}{4},3),(\frac{\pi}{2},0),(\frac{3\pi}{4}, - 3),(\pi,0)$ and connect them with a smooth curve for one - period of $y = 3\sin(2x)$.