5. challenge: a ferris wheel 72 feet in diameter makes one revolution every 122 sec. the center of the wheel…

Explanation:

Step1: Determine the amplitude

The diameter of the Ferris - wheel is 72 feet, so the amplitude $A=\frac{72}{2}=36$ (amplitude is half of the vertical distance between the maximum and minimum values).

Step2: Determine the vertical shift

The center of the wheel is 50 feet above the ground, so the vertical shift $D = 50$.

Step3: Determine the period and the value of $B$

The period $T = 122$ seconds. The formula for the period of a cosine function is $T=\frac{2\pi}{B}$. Solving for $B$, we get $B=\frac{2\pi}{T}=\frac{2\pi}{122}=\frac{\pi}{61}$.

Step4: Write the cosine function

Assuming no horizontal shift ($C = 0$), the cosine function is $y=A\cos(Bx)+D$. Substituting the values of $A$, $B$, and $D$, we have $y = 36\cos(\frac{\pi}{61}x)+50$.

To graph the function $y = 36\cos(\frac{\pi}{61}x)+50$ over the interval $[0,122]$:

• When $x = 0$, $y=36\cos(0)+50=36\times1 + 50=86$.
• When $x = 61$, $y=36\cos(\pi)+50=36\times(- 1)+50 = 14$.
• When $x = 122$, $y=36\cos(2\pi)+50=36\times1+50 = 86$.

Answer:

The cosine function is $y = 36\cos(\frac{\pi}{61}x)+50$ and to graph it, plot the points $(0,86)$, $(61,14)$, $(122,86)$ and connect them with a smooth cosine - shaped curve over the interval $[0,122]$.