Question

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

Explanation:

Step1: Find x - intercepts

Set $y = 0$, so $\frac{1}{2}\sin x=0$. Then $\sin x = 0$. In one - cycle ($0\leq x\leq2\pi$), $x = 0,\pi,2\pi$. The points are $(0,0),(\pi,0),(2\pi,0)$.

Step2: Find maxima

The maximum value of $\sin x$ is 1. For $y=\frac{1}{2}\sin x$, when $\sin x = 1$, $x=\frac{\pi}{2}$, and $y=\frac{1}{2}$. The point is $(\frac{\pi}{2},\frac{1}{2})$.

Step3: Find minima

The minimum value of $\sin x$ is - 1. For $y = \frac{1}{2}\sin x$, when $\sin x=-1$, $x=\frac{3\pi}{2}$, and $y=-\frac{1}{2}$. The point is $(\frac{3\pi}{2},-\frac{1}{2})$.

Answer:

The x - intercepts are $(0,0),(\pi,0),(2\pi,0)$; the maximum point is $(\frac{\pi}{2},\frac{1}{2})$; the minimum point is $(\frac{3\pi}{2},-\frac{1}{2})$