graph the trigonometric function.\n\n$y = - 2cos(\frac{1}{2}x)$\n\nplot all points corresponding to x…

Explanation:

Step1: Find the period

The general form of a cosine function is $y = A\cos(Bx - C)+D$. For $y=-2\cos(\frac{1}{2}x)$, $B = \frac{1}{2}$. The period $T$ of a cosine - function is given by $T=\frac{2\pi}{|B|}$. So, $T=\frac{2\pi}{\frac{1}{2}} = 4\pi$.

Step2: Find the maxima

The amplitude $A=-2$. The maximum value of $\cos(\frac{1}{2}x)$ is 1. When $\cos(\frac{1}{2}x)=1$, $y=-2\times1=-2$. Solving $\frac{1}{2}x = 2k\pi$ ($k\in\mathbb{Z}$) for $x$ in one - cycle ($0\leq x\leq4\pi$), when $k = 0$, $x = 0$; when $k = 1$, $x = 4\pi$. The points are $(0, - 2)$ and $(4\pi,-2)$.

Step3: Find the minima

The minimum value of $\cos(\frac{1}{2}x)$ is - 1. When $\cos(\frac{1}{2}x)=-1$, $y=-2\times(-1)=2$. Solving $\frac{1}{2}x=(2k + 1)\pi$ ($k\in\mathbb{Z}$) for $x$ in one - cycle ($0\leq x\leq4\pi$), when $k = 0$, $x = 2\pi$. The point is $(2\pi,2)$.

Step4: Find the x - intercepts

Set $y = 0$, then $-2\cos(\frac{1}{2}x)=0$, so $\cos(\frac{1}{2}x)=0$. Solving $\frac{1}{2}x=(2k + 1)\frac{\pi}{2}$ ($k\in\mathbb{Z}$) for $x$ in one - cycle ($0\leq x\leq4\pi$), when $k = 0$, $x=\pi$; when $k = 1$, $x = 3\pi$. The points are $(\pi,0)$ and $(3\pi,0)$.

Answer:

Plot the points $(0, - 2)$, $(2\pi,2)$, $(4\pi,-2)$, $(\pi,0)$, $(3\pi,0)$ on the given graph.