chapter 12 review question 10 of 10 (1 point) | question attempt: 1 of unlimited graph the trigonometric…

Explanation:

Step1: Recall general cosine - function form

The general form of a cosine function is $y = A\cos(Bx - C)+D$. For $y = 3\cos(x+\frac{\pi}{2})$, we have $A = 3$, $B = 1$, $C=-\frac{\pi}{2}$, $D = 0$. The period $T=\frac{2\pi}{B}=2\pi$.

Step2: Find x - intercepts

Set $y = 0$, so $3\cos(x+\frac{\pi}{2})=0$. Then $\cos(x+\frac{\pi}{2}) = 0$. We know that $\cos\theta=0$ when $\theta=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. So $x+\frac{\pi}{2}=\frac{\pi}{2}+k\pi$, which gives $x = k\pi$. For one - cycle ($0\leq x\leq2\pi$), when $k = 0,x = 0$; when $k = 1,x=\pi$; when $k = 2,x = 2\pi$.

Step3: Find maxima

The maximum value of $\cos\theta$ is 1. Set $x+\frac{\pi}{2}=2k\pi$. Then $x=2k\pi-\frac{\pi}{2}$. For one - cycle ($0\leq x\leq2\pi$), when $k = 1,x=\frac{3\pi}{2}$, and $y_{max}=3$.

Step4: Find minima

The minimum value of $\cos\theta$ is - 1. Set $x+\frac{\pi}{2}=(2k + 1)\pi$. Then $x=(2k + 1)\pi-\frac{\pi}{2}=2k\pi+\frac{\pi}{2}$. For one - cycle ($0\leq x\leq2\pi$), when $k = 0,x=\frac{\pi}{2}$, and $y_{min}=-3$.

The points to plot for one - cycle ($0\leq x\leq2\pi$) are:

• x - intercepts: $(0,0),(\pi,0),(2\pi,0)$
• maxima: $(\frac{3\pi}{2},3)$
• minima: $(\frac{\pi}{2},-3)$

Answer:

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