The following information will be used by both problem 1 and problem 2.
The waterpark owners want to know how the sales of season passes compares to the single
day passes sold throughout the season. The number of season passes sold each day of the
season is modeled by the equation 𝑦 = −
1/25x^2 + 2𝑥 + 250 while the number of single day
passes is modeled by 𝑦 = -1/40x^2+20x+200

a. How many passes were sold on day 𝑥 = 0.
b. On which day of the season did the number of single day passes equal the number of
season passes sold?

Given the equations for the number of passes sold each day, we can start solving the problems step by step.

### Problem 1: Number of Passes Sold on Day \( x = 0 \)
1. **Season Passes:**

The equation for the number of season passes sold each day is:
\[ y = -\frac{1}{25}x^2 + 2x + 250 \]

Substitute \( x = 0 \):
\[ y = -\frac{1}{25}(0)^2 + 2(0) + 250 \]
\[ y = 250 \]

So, on day \( x = 0 \), 250 season passes were sold.

2. **Single Day Passes:**

The equation for the number of single day passes sold each day is:
\[ y = -\frac{1}{40}x^2 + 20x + 200 \]

Substitute \( x = 0 \):
\[ y = -\frac{1}{40}(0)^2 + 20(0) + 200 \]
\[ y = 200 \]

So, on day \( x = 0 \), 200 single day passes were sold.

### Problem 2: Day When Single Day Passes Equal Season Passes
To find the day when the number of single day passes equals the number of season passes, we need to set the two equations equal to each other and solve for \( x \):

\[ -\frac{1}{25}x^2 + 2x + 250 = -\frac{1}{40}x^2 + 20x + 200 \]

First, clear the fractions by multiplying through by 200 (the least common multiple of 25 and 40):

\[ -8x^2 + 400x + 50000 = -5x^2 + 4000x + 40000 \]

Combine like terms:

\[ -8x^2 + 400x + 50000 = -5x^2 + 4000x + 40000 \]
\[ -8x^2 + 400x + 50000 + 5x^2 - 4000x - 40000 = 0 \]
\[ -3x^2 - 3600x + 10000 = 0 \]

This is a quadratic equation, which can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = -3 \), \( b = -3600 \), and \( c = 10000 \):

\[ x = \frac{-(-3600) \pm \sqrt{(-3600)^2 - 4(-3)(10000)}}{2(-3)} \]
\[ x = \frac{3600 \pm \sqrt{12960000 + 120000}}{-6} \]
\[ x = \frac{3600 \pm \sqrt{13080000}}{-6} \]
\[ x = \frac{3600 \pm 3616.36}{-6} \]

This gives us two potential solutions for \( x \):

\[ x = \frac{3600 + 3616.36}{-6} = \frac{7216.36}{-6} = -1202.73 \] (not a feasible solution since \( x \) must be a positive day of the season)
\[ x = \frac{3600 - 3616.36}{-6} = \frac{-16.36}{-6} = 2.73 \]

Since the day must be an integer, we round to the nearest whole number, so the number of single day passes equals the number of season passes sold on day \( x = 3 \).

### Summary
a. On day \( x = 0 \):
- 250 season passes were sold.
- 200 single day passes were sold.

b. The number of single day passes equals the number of season passes on day \( x = 3 \).

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How do you derive the quadratic formula?
2. How can we graph the equations for visual comparison?
3. What is the significance of the coefficients in these equations?
4. How do we interpret the vertex of these parabolas?
5. What other methods can we use to solve quadratic equations?

**Tip:** When solving quadratic equations, always check if the solutions make sense in the context of the problem, such as ensuring the days are positive integers.

Explore how to compare sales of season passes and single day passes using quadratic equations. Learn how to calculate the number of passes sold on day x = 0 and determine the day when single day passes equal season passes. Suitable for high school students studying quadratic equations.

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