The image shows a challenge asking to plot a parabola through three given points:

• $(-9, 4)$
• $(-6, 1)$
• $(-3, 4)$

To find the equation of the parabola that passes through these points, we can assume a general quadratic equation:

$y = ax^2 + bx + c$

We substitute the three points into this equation to form a system of three equations to solve for $a$, $b$, and $c$.

1. For $(-9, 4)$: $4 = a(-9)^2 + b(-9) + c$ $4 = 81a - 9b + c$

2. For $(-6, 1)$: $1 = a(-6)^2 + b(-6) + c$ $1 = 36a - 6b + c$

3. For $(-3, 4)$: $4 = a(-3)^2 + b(-3) + c$ $4 = 9a - 3b + c$

Now, I will solve this system of equations to find $a$, $b$, and $c$.The solution to the system of equations is:

• $a = \frac{1}{3}$
• $b = 4$
• $c = 13$

Thus, the equation of the parabola passing through the points $(-9, 4)$, $(-6, 1)$, and $(-3, 4)$ is:

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

Would you like further explanation on solving these types of systems, or any additional details?

Here are 5 related questions to consider:

1. How can you verify if a quadratic equation fits multiple points?
2. What happens to the parabola if you change the sign of $a$?
3. Can you generalize the process of finding a parabola for any set of three points?
4. How does the vertex form of a parabola relate to its standard form?
5. How would you solve this system graphically instead of algebraically?

Tip: When working with systems of equations, always check for consistency by substituting the solutions back into the original equations.