To find the correct sine function that models the attendance at the state park, let's follow a detailed, step-by-step process.
1. Identify the amplitude:
- The attendance varies from a low of [tex]$1,000,000$[/tex] visitors in September to a high of [tex]$2,000,000$[/tex] visitors in March.
- The amplitude of our sine function is half the difference between the maximum and minimum values.
- [tex]\[ \text{Amplitude} = \frac{\text{high} - \text{low}}{2} = \frac{2,000,000 - 1,000,000}{2} = 0.5 \text{ million visitors} \][/tex]
2. Determine the midline:
- The midline is the average value of the maximum and minimum attendance.
- [tex]\[ \text{Midline} = \frac{\text{high} + \text{low}}{2} = \frac{2,000,000 + 1,000,000}{2} = 1.5 \text{ million visitors} \][/tex]
3. Period of the sine function:
- The attendance cycle completes in one year, indicating a period of 12 months.
- In general, a sine function [tex]\(f(t) = A \sin(B t + C) + D\)[/tex] has a period of [tex]\(\frac{2\pi}{B}\)[/tex].
- Since the period is 12 months, [tex]\(B\)[/tex] must be:
- [tex]\[ 12 = \frac{2\pi}{B} \Rightarrow B = \frac{\pi}{6} \][/tex]
4. Determine the phase shift:
- We need to align our sine function with the data points. The high occurs in March (t=3) and the low in September (t=9).
- For simplifying, let's first determine the phase shift needed for March.
- A sine function peaks at [tex]\(\frac{\pi}{2}\)[/tex] for a standard [tex]\(\sin(x)\)[/tex].
- Since the high is at [tex]\(t=3\)[/tex], we need:
- [tex]\[ \frac{\pi}{6} \cdot 3 + \phi = \frac{\pi}{2} \Rightarrow \phi = \frac{\pi}{2} - \frac{3\pi}{6} = \frac{\pi}{2} - \frac{\pi}{2} = 0 \][/tex]
With the parameters calculated, let’s form our sine equation with those parameters:
[tex]\[ N(t) = 0.5 \sin \left( \frac{\pi}{6} t - \frac{\pi}{3} \right) + 1.5 \][/tex]
The sine function options we have are:
- [tex]\[ N(t) = 1.5 \sin \left(\frac{\pi}{6} t\right) + 0.5 \][/tex]
- [tex]\[ N(t) = 2 \sin \left(\frac{\pi}{6} t\right) - 1 \][/tex]
- [tex]\[ N(t) = 0.5 \sin (2 \pi t) + 1.5 \][/tex]
- [tex]\[ N(t) = 0.5 \sin \left(\frac{\pi}{6} t\right) + 1.5 \][/tex]
After careful examination of our parameters:
- Correct amplitude: 0.5
- Correct midline: 1.5
- Correct period: [tex]\( \frac{\pi}{6} \)[/tex]
Thus, the function that correctly models the given behavior is:
[tex]\[
\boxed{0.5 \sin \left( \frac{\pi}{6} t \right) + 1.5}
\][/tex]
