graph the trigonometric function.\ny = 2cos3x\nplot all points corresponding to x - intercepts, minima, and…

Explanation:

Step1: Find the period

For $y = A\cos(Bx)$, period $T=\frac{2\pi}{B}$. Here $B = 3$, so $T=\frac{2\pi}{3}$.

Step2: Find the maxima

The amplitude $A = 2$. The maximum value of $\cos(3x)$ is 1. So $y_{max}=2\times1 = 2$. When $\cos(3x)=1$, $3x = 2k\pi$, $x=\frac{2k\pi}{3}, k\in\mathbb{Z}$. In one - cycle $[0,\frac{2\pi}{3}]$, when $k = 0$, $x = 0$, the maximum point is $(0,2)$.

Step3: Find the minima

The minimum value of $\cos(3x)$ is - 1. So $y_{min}=2\times(-1)=-2$. When $\cos(3x)=-1$, $3x=(2k + 1)\pi$, $x=\frac{(2k + 1)\pi}{3},k\in\mathbb{Z}$. In one - cycle $[0,\frac{2\pi}{3}]$, when $k = 0$, $x=\frac{\pi}{3}$, the minimum point is $(\frac{\pi}{3},-2)$.

Step4: Find the x - intercepts

Set $y = 0$, then $2\cos(3x)=0$, so $\cos(3x)=0$. Then $3x=(2k + 1)\frac{\pi}{2}$, $x=(2k + 1)\frac{\pi}{6},k\in\mathbb{Z}$. In one - cycle $[0,\frac{2\pi}{3}]$, when $k = 0$, $x=\frac{\pi}{6}$ and when $k = 1$, $x=\frac{\pi}{2}$. The x - intercepts are $(\frac{\pi}{6},0)$ and $(\frac{\pi}{2},0)$.

Answer:

Plot the points $(0,2),(\frac{\pi}{6},0),(\frac{\pi}{3},-2),(\frac{\pi}{2},0)$ within the cycle $[0,\frac{2\pi}{3}]$ on the graph.