Question

a passenger is riding a ferris wheel. the graph shows the height, in feet, of the passenger as a function of time.
move values to the boxes to create a function, h(t), that models the height, in feet, of the passenger after t minutes on the ferris wheel.
h(t) = - square cos( square pi t) + square
\frac{1}{3}
3
5
20
25
40

Explanation:

Step1: Determine the amplitude

The amplitude \( A \) of a cosine function \( y = A\cos(Bx)+C \) (here we have a negative sign, so \( -A\cos(Bx)+C \)) is half the distance between the maximum and minimum values. From the graph, the maximum height is 45 feet and the minimum is 5 feet. So the amplitude \( A=\frac{45 - 5}{2}=\frac{40}{2} = 20\). But since the function is \( h(t)=-\square\cos(\square\pi t)+\square \), the first box (the coefficient of cosine) will be 20.

Step2: Determine the period and \( B \)

The period \( T \) of the cosine function is the time it takes to complete one full cycle. From the graph, we can see that the period \( T=\frac{2}{3} \) minutes? Wait, no, looking at the x - axis (time in minutes), from 0 to \( \frac{2}{3} \)? Wait, no, let's check the number of cycles. Wait, the general formula for the period of \( y = \cos(Bx) \) is \( T=\frac{2\pi}{B} \). Looking at the graph, in 2 minutes, how many cycles? Wait, from the graph, let's see the period. Wait, the standard form for a cosine function with period \( T \) is \( B=\frac{2\pi}{T} \). But in our function, the argument is \( \square\pi t \), so \( B\pi \). Let's find the period. Let's see, when \( t = 0 \), what's the value? If we use the cosine function, \( \cos(0)=1 \), so \( h(0)=-A\cos(0)+C=-A + C \). At \( t = 0 \), the height should be the minimum? Wait, no, the function is \( h(t)=-A\cos(B\pi t)+C \). When \( t = 0 \), \( \cos(0)=1 \), so \( h(0)=-A + C \). The minimum height is 5, and the maximum is 45. So when \( \cos(B\pi t)= - 1 \), \( h(t)=A + C \) (maximum), and when \( \cos(B\pi t)=1 \), \( h(t)=-A + C \) (minimum). So we have:
\( A + C=45 \) and \( -A + C = 5 \)
Solving these two equations: add them together: \( (A + C)+(-A + C)=45 + 5\Rightarrow2C = 50\Rightarrow C = 25 \). Then from \( -A + 25=5\Rightarrow A = 20 \), which matches our earlier amplitude calculation.
Now, let's find \( B \). The period \( T \): let's find the time it takes to go from maximum to minimum to maximum. Wait, the period \( T \) is the time for one full cycle. Let's use the two - point form. We know that \( C = 25 \) (the vertical shift, the midline, which is \( \frac{45 + 5}{2}=25 \)). Now, the period \( T \): from the graph, let's see how long it takes to complete one cycle. Let's take two consecutive peaks or troughs. Wait, the function is \( h(t)=-20\cos(B\pi t)+25 \). Let's find \( B \). The period \( T=\frac{2\pi}{B\pi}=\frac{2}{B} \) (since the argument is \( B\pi t \), so the period formula is \( T=\frac{2\pi}{B\pi}=\frac{2}{B} \)). From the graph, let's see the period. Let's look at the time when the function completes one cycle. Let's take \( t = 0 \), \( h(0)=-20\cos(0)+25=-20 + 25 = 5 \) (which is the minimum, that makes sense because we have a negative cosine, so it's a reflection over the x - axis, so at \( t = 0 \), \( \cos(0)=1 \), so \( - 20\times1+25 = 5 \), which is the minimum). Then when does it reach the maximum? The maximum occurs when \( \cos(B\pi t)=-1 \), so \( h(t)=20 + 25=45 \). The time to go from \( t = 0 \) (minimum) to the next maximum: the period is the time between two consecutive minima or two consecutive maxima. Let's see, in 2 minutes, how many cycles? Wait, no, let's use the period formula. We know that \( C = 25 \), \( A = 20 \). Now, let's find \( B \). Let's look at the period. Let's see, when \( t=\frac{1}{3} \), what happens? Wait, the function is \( h(t)=-20\cos(B\pi t)+25 \). Let's find \( B \) such that the period is \( \frac{2}{3} \)? Wait, no, let's use the fact that in the function \( h(t)=-A\cos(B\pi t)+C \), the period \( T=\frac{2}{B} \) (since \( \omega = B\pi \), and \( T=\frac{2\pi}{\omega}=\frac{2\pi}{B\pi}=\frac{2}{B} \)). From the graph, let's count the number of cycles in 2 minutes. Wait, the graph shows that in 2 minutes, there are 3 cycles? Wait, no, looking at the x - axis (time in minutes), from 0 to 2, how many peaks? Let's see, the peaks are at certain times. Wait, the correct way: we know that \( C = 25 \) (midline), \( A = 20 \) (amplitude). Now, let's find \( B \). Let's take the period. Let's suppose that in \( t = \frac{2}{3} \) minutes, one cycle? No, wait, the function is \( h(t)=-20\cos(3\pi t)+25 \). Let's check the period: \( T=\frac{2\pi}{3\pi}=\frac{2}{3} \) minutes. Let's check at \( t = 0 \), \( h(0)=-20\cos(0)+25=-20 + 25 = 5 \) (minimum). At \( t=\frac{1}{6} \), \( 3\pi\times\frac{1}{6}=\frac{\pi}{2} \), \( \cos(\frac{\pi}{2}) = 0 \), \( h(\frac{1}{6})=-20\times0 + 25 = 25 \) (midline). At \( t=\frac{1}{3} \), \( 3\pi\times\frac{1}{3}=\pi \), \( \cos(\pi)=-1 \), \( h(\frac{1}{3})=-20\times(-1)+25=20 + 25 = 45 \) (maximum). Then at \( t=\frac{2}{3} \), \( 3\pi\times\frac{2}{3}=2\pi \), \( \cos(2\pi)=1 \), \( h(\frac{2}{3})=-20\times1+25 = 5 \) (minimum), which matches the period of \( \frac{2}{3} \) minutes. So \( B = 3 \), so the second box (the coefficient of \( \pi t \)) is 3.

Step3: Determine the vertical shift \( C \)

The vertical shift \( C \) is the midline, which is the average of the maximum and minimum values. So \( C=\frac{45 + 5}{2}=\frac{50}{2}=25 \). So the third box is 25.

Answer:

\( h(t)=-20\cos(3\pi t)+25 \) So the boxes are filled as 20, 3, and 25.