Question

graph $h(x) = 7cosleft(\frac{1}{2}x
ight) + 1$.
use 3.14 for $\pi$.
use the sine tool to graph the function. the first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

Explanation:

Step1: Identify midline

The midline of $h(x)=A\cos(Bx)+k$ is $y=k$. Here $k=1$, so midline is $y=1$.

Step2: Find x on midline

For cosine function, midline points occur when $\cos(Bx)=0$. Let $Bx=\frac{\pi}{2}$, $B=\frac{1}{2}$, so $x=\pi\approx3.14$? Wait no—wait, initial midline point: when $x=0$, $\cos(0)=1$, not 0. Wait, midline when $\cos(\frac{1}{2}x)=0$: $\frac{1}{2}x=\frac{\pi}{2}\Rightarrow x=\pi\approx3.14$, then $h(3.14)=7*0+1=1$, so first point $(3.14,1)$.

Step3: Closest max/min

Closest to $x=3.14$: max when $\cos(\frac{1}{2}x)=1\Rightarrow \frac{1}{2}x=0\Rightarrow x=0$, $h(0)=7*1+1=8$ (max) or min when $\cos(\frac{1}{2}x)=-1\Rightarrow x=2\pi\approx6.28$, $h(6.28)=7*(-1)+1=-6$. Distance from $3.14$: $|0-3.14|=3.14$, $|6.28-3.14|=3.14$. Either works, but max is $(0,8)$.

Corrected: First midline point $(3.14,1)$, second closest max $(0,8)$ or min $(6.28,-6)$.

Final answer: First point $(3.14,1)$, second point $(0,8)$ (or $(6.28,-6)$)

Answer:

First point: $(0, 1)$; Second point: $(0, 8)$