Question

graph the trigonometric function. (y = \frac{1}{2}sin(\frac{2}{3}x)) plot all points corresponding to x - intercepts, minima, and maxima within one cycle. then click on the graph - a - function button.

Explanation:

Step1: Recall the general form of sine - function

The general form of a sine - function is $y = A\sin(Bx - C)+D$. For the function $y=\frac{1}{2}\sin(\frac{2}{3}x)$, we have $A=\frac{1}{2}$, $B = \frac{2}{3}$, $C = 0$, and $D = 0$.

Step2: Find the period

The period of the sine - function $y = A\sin(Bx - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Substituting $B=\frac{2}{3}$ into the formula, we get $T=\frac{2\pi}{\frac{2}{3}}=3\pi$.

Step3: Find the x - intercepts

Set $y = 0$. Then $\frac{1}{2}\sin(\frac{2}{3}x)=0$, which implies $\sin(\frac{2}{3}x)=0$. So, $\frac{2}{3}x = k\pi$, where $k\in\mathbb{Z}$. Solving for $x$, we have $x=\frac{3}{2}k\pi$. For one - cycle ($0\leq x\leq3\pi$), when $k = 0$, $x = 0$; when $k = 1$, $x=\frac{3\pi}{2}$; when $k = 2$, $x = 3\pi$.

Step4: Find the maxima and minima

The amplitude of the function is $|A|=\frac{1}{2}$. The maximum value of $y=\frac{1}{2}\sin(\frac{2}{3}x)$ occurs when $\sin(\frac{2}{3}x)=1$. Then $\frac{2}{3}x=\frac{\pi}{2}+2k\pi$, and $x=\frac{3\pi}{4}+3k\pi$. In the interval $[0,3\pi]$, when $k = 0$, $x=\frac{3\pi}{4}$ and $y=\frac{1}{2}$. The minimum value of $y=\frac{1}{2}\sin(\frac{2}{3}x)$ occurs when $\sin(\frac{2}{3}x)=-1$. Then $\frac{2}{3}x=\frac{3\pi}{2}+2k\pi$, and $x=\frac{9\pi}{4}+3k\pi$. In the interval $[0,3\pi]$, when $k = 0$, $x=\frac{9\pi}{4}$ and $y=-\frac{1}{2}$.

To graph the function, plot the x - intercepts $(0,0),(\frac{3\pi}{2},0),(3\pi,0)$, the maximum point $(\frac{3\pi}{4},\frac{1}{2})$, and the minimum point $(\frac{9\pi}{4},-\frac{1}{2})$ and then connect these points with a smooth curve.

Answer:

Plot the points $(0,0),(\frac{3\pi}{2},0),(3\pi,0),(\frac{3\pi}{4},\frac{1}{2}),(\frac{9\pi}{4},-\frac{1}{2})$ and connect them with a smooth curve for one - cycle of the function $y=\frac{1}{2}\sin(\frac{2}{3}x)$.