Let's address each part of the problem step-by-step.
### Part (a): Writing Functions for Each Option
Option A: Charge a \[tex]$10 entrance fee and sell game tickets for \$[/tex]0.25 each.
- Let [tex]\( x \)[/tex] be the number of game tickets sold.
- The total profit [tex]\( P_A \)[/tex] can be modeled as:
[tex]\[ P_A(x) = 10 + 0.25x \][/tex]
Option B: Charge \[tex]$0.50 per game ticket without an entrance fee.
- Let \( x \) be the number of game tickets sold.
- The total profit \( P_B \) can be modeled as:
\[ P_B(x) = 0.50x \]
### Part (b): Finding the Intersection Points Algebraically
We are given two equations:
1. \( y = x^2 + 3x - 4 \)
2. \( y = -x - 7 \)
To find the intersection, set the equations equal to each other:
\[ x^2 + 3x - 4 = -x - 7 \]
Step 1: Move all terms to one side:
\[ x^2 + 3x - 4 + x + 7 = 0 \]
\[ x^2 + 4x + 3 = 0 \]
Step 2: Factor the quadratic equation:
\[ (x + 1)(x + 3) = 0 \]
Step 3: Solve for \( x \):
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
Step 4: Find the corresponding \( y \) values by substituting \( x \) back into either equation (we'll use \( y = -x - 7 \)):
For \( x = -1 \):
\[ y = -(-1) - 7 = 1 - 7 = -6 \]
For \( x = -3 \):
\[ y = -(-3) - 7 = 3 - 7 = -4 \]
Thus, the intersection points are:
\[ (-1, -6) \quad \text{and} \quad (-3, -4) \]
Next, we solve the second equation:
\[ 3|x + 4| - 2 = 4 \]
Step 1: Isolate the absolute value term:
\[ 3|x + 4| - 2 = 4 \]
\[ 3|x + 4| = 6 \]
\[ |x + 4| = 2 \]
Step 2: Solve the absolute value equation:
\[ x + 4 = 2 \quad \Rightarrow \quad x = -2 \]
\[ x + 4 = -2 \quad \Rightarrow \quad x = -6 \]
### Intersection Points from Part (b)
- The intersections of \( y = x^2 + 3x - 4 \) and \( y = -x - 7 \) are \( (-1, -6) \) and \( (-3, -4) \).
- The solutions for \( 3|x + 4| - 2 = 4 \) are \( x = -2 \) and \( x = -6 \).
### Part (c): Advice Based on the Intersection Points
Based on the intersection points, we can determine that \( x = -2 \) and \( x = -6 \) are significant values where the functions of ticket sales equate to each other.
However, since ticket numbers cannot be negative in a realistic scenario, these solutions do not directly help in deciding which option to choose.
In terms of maximizing profits:
- For Option A: Profit will always include the fixed \$[/tex]10 plus [tex]\( 0.25x \)[/tex] per ticket.
- For Option B: Profit will be [tex]\( 0.50x \)[/tex] per ticket.
To truly maximize profits, consider the potential number of tickets sold:
- If fewer tickets are expected to be sold, Option A might be preferable due to the fixed entrance fee, ensuring at least \$10.
- If a high number of tickets are expected to be sold, Option B might yield higher profits due to the higher per-ticket charge.
By factoring any assumptions about the number of tickets, your friend can choose the best option for maximizing the fundraiser's profits.
