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The challenge asks you to plot a parabola that passes through three points: \((-9, 4)\), \((-3, 4)\), and \((-6, 1)\).

To find the equation of the parabola, let's assume the general form of a parabola as \( y = ax^2 + bx + c \). We can substitute the three given points into this equation to create a system of three equations:

1. Substituting \((-9, 4)\): 
   \[ 4 = a(-9)^2 + b(-9) + c \Rightarrow 4 = 81a - 9b + c \]

2. Substituting \((-3, 4)\): 
   \[ 4 = a(-3)^2 + b(-3) + c \Rightarrow 4 = 9a - 3b + c \]

3. Substituting \((-6, 1)\): 
   \[ 1 = a(-6)^2 + b(-6) + c \Rightarrow 1 = 36a - 6b + c \]

Now we have the system of equations:
\[
\begin{aligned}
81a - 9b + c &= 4 \\
9a - 3b + c &= 4 \\
36a - 6b + c &= 1
\end{aligned}
\]

I'll solve this system of equations to find the values of \(a\), \(b\), and \(c\).The equation of the parabola that passes through the points \((-9, 4)\), \((-3, 4)\), and \((-6, 1)\) is:

\[
y = \frac{1}{3}x^2 + 4x + 13
\]

This is one possible parabola equation.

Would you like to explore another method or approach for the second parabola? Let me know if you'd like further details or have any questions!

Here are some related questions:
1. How can we find the vertex form of this parabola?
2. Can we verify the result by plotting the points and equation?
3. What are the x-intercepts of this parabola, if any?
4. How does changing the points affect the equation of the parabola?
5. Can we convert this standard form equation into vertex form?

**Tip:** When given three points, always check if two y-values are the same—this often simplifies solving for the parabola.

Plot a parabola through the points (-9, 4), (-6, 1), and (-3, 4).

Learn how to plot a quadratic parabola through the points (-9, 4), (-6, 1), and (-3, 4) by solving a system of equations. This involves algebraic manipulation and graphing, suitable for Grades 9-11.

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