plot 5 points representing one full cycle of the function.\n\n$f(x)=6cos(\frac{pi}{2}(x + 3))+6$

Explanation:

Step1: Find the period

The general form of a cosine function is $y = A\cos(B(x - C))+D$. For $f(x)=6\cos(\frac{\pi}{2}(x + 3))+6$, $B=\frac{\pi}{2}$. The period $T=\frac{2\pi}{B}=\frac{2\pi}{\frac{\pi}{2}} = 4$.

Step2: Determine key - points

We know that for $y = \cos(x)$, the key - points in one period $[0,2\pi]$ are $x = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$. For $y = 6\cos(\frac{\pi}{2}(x + 3))+6$, we set $\frac{\pi}{2}(x + 3)$ equal to these values and solve for $x$.

When $\frac{\pi}{2}(x + 3)=0$

Solve for $x$: $\frac{\pi}{2}(x + 3)=0\Rightarrow x+3 = 0\Rightarrow x=-3$. Then $f(-3)=6\cos(0)+6=6\times1 + 6=12$.

When $\frac{\pi}{2}(x + 3)=\frac{\pi}{2}$

Solve for $x$: $\frac{\pi}{2}(x + 3)=\frac{\pi}{2}\Rightarrow x + 3=1\Rightarrow x=-2$. Then $f(-2)=6\cos(\frac{\pi}{2})+6=6\times0 + 6=6$.

When $\frac{\pi}{2}(x + 3)=\pi$

Solve for $x$: $\frac{\pi}{2}(x + 3)=\pi\Rightarrow x + 3 = 2\Rightarrow x=-1$. Then $f(-1)=6\cos(\pi)+6=6\times(-1)+6 = 0$.

When $\frac{\pi}{2}(x + 3)=\frac{3\pi}{2}$

Solve for $x$: $\frac{\pi}{2}(x + 3)=\frac{3\pi}{2}\Rightarrow x + 3=3\Rightarrow x=0$. Then $f(0)=6\cos(\frac{3\pi}{2})+6=6\times0 + 6=6$.

When $\frac{\pi}{2}(x + 3)=2\pi$

Solve for $x$: $\frac{\pi}{2}(x + 3)=2\pi\Rightarrow x + 3 = 4\Rightarrow x=1$. Then $f(1)=6\cos(2\pi)+6=6\times1+6 = 12$.

The five points are $(-3,12),(-2,6),(-1,0),(0,6),(1,12)$.

Answer:

$(-3,12),(-2,6),(-1,0),(0,6),(1,12)$