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Let's break down the problem step-by-step to solve it.

**Problem Statement:**
You need to create a route for Tony Hawk from point \( T (-4, 1) \) to point \( P (14, 16) \) while avoiding the garbage cans at points \( A (-3, 2) \), \( B (8, 6) \), \( C (0, 5) \), \( D (4, 12) \), \( E (14, 13) \), \( F (11, 15) \), and \( G (3, 7) \). The route will be a combination of linear and quadratic equations.

**Requirements:**
1. At least one parabola that opens up.
2. At least one parabola that opens down.
3. At least one of the parabolas must have a stretch factor.
4. At least two lines.
5. Key points should be clearly labeled as ordered pairs and lie on major gridlines (no uneven decimals).
6. Four relations (equations) should be labeled.

**Steps to solve:**

### 1. Graphing the Path

- **Parabola 1 (Opens Up):** We can place a parabola that opens up with vertex at \( V_1 = (2, 4) \).
- **Parabola 2 (Opens Down):** We can place a parabola that opens down with vertex at \( V_2 = (10, 10) \).

- **Line 1:** We can use a line passing through \( T (-4, 1) \) and another point avoiding the garbage cans, say point \( (0, 3) \).
- **Line 2:** We can use a line passing through \( P (14, 16) \) and another point avoiding the garbage cans, say point \( (12, 14) \).

### 2. Equation of the Path

**Parabola 1 (Opens Up):**
\[ y = a(x - h)^2 + k \]
For the vertex at \( (2, 4) \):
\[ y = a(x - 2)^2 + 4 \]
Choose a point on the parabola to find \( a \), for example, \( (0, 0) \):
\[ 0 = a(0 - 2)^2 + 4 \]
\[ 0 = 4a + 4 \]
\[ a = -1 \]
So the equation is:
\[ y = -(x - 2)^2 + 4 \]

**Parabola 2 (Opens Down):**
\[ y = b(x - h)^2 + k \]
For the vertex at \( (10, 10) \):
\[ y = b(x - 10)^2 + 10 \]
Choose a point on the parabola to find \( b \), for example, \( (12, 8) \):
\[ 8 = b(12 - 10)^2 + 10 \]
\[ 8 = 4b + 10 \]
\[ b = -\frac{1}{2} \]
So the equation is:
\[ y = -\frac{1}{2}(x - 10)^2 + 10 \]

**Line 1:**
\[ y = mx + c \]
Passing through \( (-4, 1) \) and \( (0, 3) \):
\[ m = \frac{3 - 1}{0 + 4} = \frac{1}{2} \]
\[ y = \frac{1}{2}x + c \]
Substituting \( (-4, 1) \):
\[ 1 = \frac{1}{2}(-4) + c \]
\[ 1 = -2 + c \]
\[ c = 3 \]
So the equation is:
\[ y = \frac{1}{2}x + 3 \]

**Line 2:**
\[ y = nx + d \]
Passing through \( (14, 16) \) and \( (12, 14) \):
\[ n = \frac{16 - 14}{14 - 12} = 1 \]
\[ y = x + d \]
Substituting \( (12, 14) \):
\[ 14 = 12 + d \]
\[ d = 2 \]
So the equation is:
\[ y = x + 2 \]

### Summary of Equations:
1. \( y = -(x - 2)^2 + 4 \)
2. \( y = -\frac{1}{2}(x - 10)^2 + 10 \)
3. \( y = \frac{1}{2}x + 3 \)
4. \( y = x + 2 \)

Do you want more details or have any questions? Here are five questions you might consider asking next:

1. Can you show how to graph these equations?
2. How do you determine the stretch factor in a parabola?
3. What is the process for finding the vertex of a parabola given its equation?
4. Can you explain how to find the intersection points of these equations?
5. How do you ensure that the path avoids all the specified points (garbage cans)?

**Tip:** When working with quadratic equations, remember that the vertex form \( y = a(x - h)^2 + k \) makes it easy to identify the vertex and direction of the parabola.

Learn how to plan a route using quadratic and linear equations with a detailed step-by-step solution. This problem involves creating a path from point T (-4, 1) to point P (14, 16), avoiding specified points, and includes parabolas that open up and down, along with linear equations. Suitable for high school students.

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