graph the trigonometric function. y = 5/2 cos x plot all points corresponding to x - intercepts, minima, and…

Explanation:

Step1: Recall properties of cosine function

The general form of a cosine - function is $y = A\cos(Bx - C)+D$. For the function $y=\frac{5}{2}\cos x$, we have $A = \frac{5}{2}$, $B = 1$, $C = 0$, and $D = 0$. The period of the cosine function $y = \cos x$ is $T=\frac{2\pi}{|B|}=2\pi$.

Step2: Find the maxima

The maximum value of the cosine function $\cos x$ is 1. For $y=\frac{5}{2}\cos x$, when $\cos x = 1$, $x = 2k\pi,k\in\mathbb{Z}$. In the interval $[0,2\pi]$, when $x = 0$, $y=\frac{5}{2}\times1=\frac{5}{2}$. So the maximum - point in one cycle is $(0,\frac{5}{2})$.

Step3: Find the minima

The minimum value of the cosine function $\cos x$ is - 1. For $y=\frac{5}{2}\cos x$, when $\cos x=-1$, $x=(2k + 1)\pi,k\in\mathbb{Z}$. In the interval $[0,2\pi]$, when $x=\pi$, $y=\frac{5}{2}\times(-1)=-\frac{5}{2}$. So the minimum - point in one cycle is $(\pi,-\frac{5}{2})$.

Step4: Find the x - intercepts

Set $y = 0$, then $\frac{5}{2}\cos x=0$, which implies $\cos x = 0$. So $x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. In the interval $[0,2\pi]$, the $x$ - intercepts are $(\frac{\pi}{2},0)$ and $(\frac{3\pi}{2},0)$.

Answer:

The points to plot in one cycle ($[0,2\pi]$) are: maxima $(0,\frac{5}{2})$, minima $(\pi,-\frac{5}{2})$, and $x$ - intercepts $(\frac{\pi}{2},0)$ and $(\frac{3\pi}{2},0)$.