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To solve for the quadratic equation \( y = ax^2 + bx + c \) that goes through the points (-5, -60), (0, -5), and (13, -96), we'll use the fact that these points satisfy the quadratic equation. This gives us a system of three equations:

1. For \( (-5, -60) \):
   \[
   a(-5)^2 + b(-5) + c = -60 \quad \Rightarrow \quad 25a - 5b + c = -60
   \]

2. For \( (0, -5) \):
   \[
   a(0)^2 + b(0) + c = -5 \quad \Rightarrow \quad c = -5
   \]

3. For \( (13, -96) \):
   \[
   a(13)^2 + b(13) + c = -96 \quad \Rightarrow \quad 169a + 13b + c = -96
   \]

Now, we substitute \( c = -5 \) into the first and third equations:

1. \( 25a - 5b - 5 = -60 \quad \Rightarrow \quad 25a - 5b = -55 \)
2. \( 169a + 13b - 5 = -96 \quad \Rightarrow \quad 169a + 13b = -91 \)

We now solve the system:

\[
\begin{aligned}
25a - 5b &= -55 \quad \text{(Equation 1)} \\
169a + 13b &= -91 \quad \text{(Equation 2)}
\end{aligned}
\]

### Step 1: Solve Equation 1 for \( b \)
From \( 25a - 5b = -55 \), solve for \( b \):
\[
-5b = -55 - 25a \quad \Rightarrow \quad b = 5 + 5a
\]

### Step 2: Substitute into Equation 2
Substitute \( b = 5 + 5a \) into \( 169a + 13b = -91 \):
\[
169a + 13(5 + 5a) = -91 \quad \Rightarrow \quad 169a + 65 + 65a = -91
\]
\[
234a + 65 = -91 \quad \Rightarrow \quad 234a = -156 \quad \Rightarrow \quad a = \frac{-156}{234} = -\frac{2}{3}
\]

### Step 3: Solve for \( b \)
Substitute \( a = -\frac{2}{3} \) into \( b = 5 + 5a \):
\[
b = 5 + 5\left(-\frac{2}{3}\right) = 5 - \frac{10}{3} = \frac{15}{3} - \frac{10}{3} = \frac{5}{3}
\]

### Step 4: Write the final quadratic equation
Since we already know that \( c = -5 \), the quadratic equation is:

\[
y = -\frac{2}{3}x^2 + \frac{5}{3}x - 5
\]

Would you like more details on any of the steps, or have any specific questions?

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Here are 5 related questions to expand on this topic:
1. How would the quadratic equation change if one of the points were different?
2. What is the significance of the coefficients \( a \), \( b \), and \( c \) in the quadratic equation?
3. How can we use matrices to solve systems of equations like this one?
4. Can this quadratic equation be graphed easily, and how would you describe its key features?
5. What are some real-world applications of finding quadratic equations through points?

**Tip:** When solving systems of equations, organizing and simplifying the problem step by step can help avoid mistakes and keep your work clear.

Use quadratic regression to find the equation for the parabola going through these 3 points: (-5, -60), (0, -5), (13, -96).

This detailed solution explains how to find the quadratic equation y = ax^2 + bx + c for a parabola passing through the points (-5, -60), (0, -5), and (13, -96). Step-by-step process includes solving systems of equations and applying algebraic methods.

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