graph the trigonometric function. y = sin 3x plot all points corresponding to x - intercepts, minima, and…

Explanation:

Step1: Find the period

The general form of a sine - function is $y = A\sin(Bx - C)+D$. For $y=\sin(3x)$, $B = 3$. The period $T$ of a sine - function is given by $T=\frac{2\pi}{|B|}$. So, $T=\frac{2\pi}{3}$.

Step2: Find x - intercepts

Set $y = 0$. Then $\sin(3x)=0$. So, $3x = k\pi$, where $k\in\mathbb{Z}$. Solving for $x$, we get $x=\frac{k\pi}{3}$. For one - cycle ($0\leq x\leq\frac{2\pi}{3}$), when $k = 0$, $x = 0$; when $k = 1$, $x=\frac{\pi}{3}$; when $k = 2$, $x=\frac{2\pi}{3}$.

Step3: Find maxima and minima

The derivative of $y=\sin(3x)$ is $y'=3\cos(3x)$. Set $y' = 0$, then $\cos(3x)=0$. So, $3x=(2k + 1)\frac{\pi}{2}$, and $x=(2k + 1)\frac{\pi}{6}$. For one - cycle ($0\leq x\leq\frac{2\pi}{3}$): When $k = 0$, $x=\frac{\pi}{6}$, and $y=\sin(3\times\frac{\pi}{6})=\sin(\frac{\pi}{2}) = 1$ (maximum). When $k = 1$, $x=\frac{\pi}{2}$, and $y=\sin(3\times\frac{\pi}{2})=-1$ (minimum).

The points to plot for one - cycle ($0\leq x\leq\frac{2\pi}{3}$) are:

• $x$ - intercepts: $(0,0),(\frac{\pi}{3},0),(\frac{2\pi}{3},0)$
• Maximum: $(\frac{\pi}{6},1)$
• Minimum: $(\frac{\pi}{2},-1)$

Answer:

Plot the points $(0,0),(\frac{\pi}{6},1),(\frac{\pi}{3},0),(\frac{\pi}{2},-1),(\frac{2\pi}{3},0)$ on the given graph.