We are given the quadratic function $f(x) = ax^2 + bx + c$, and it passes through the points $(0, -8)$, $(3, 10)$, and $(6, 34)$. The question asks to find the value of the function when $x = -3$.

Step 1: Set up the system of equations

Using the points provided, we can set up three equations for $a$, $b$, and $c$.

1. At $x = 0$, $f(0) = -8$: $a(0)^2 + b(0) + c = -8 \implies c = -8$

2. At $x = 3$, $f(3) = 10$: $a(3)^2 + b(3) + c = 10 \implies 9a + 3b - 8 = 10 \implies 9a + 3b = 18$ Simplifying this equation: $3a + b = 6 \quad \text{(Equation 1)}$

3. At $x = 6$, $f(6) = 34$: $a(6)^2 + b(6) + c = 34 \implies 36a + 6b - 8 = 34 \implies 36a + 6b = 42$ Simplifying this equation: $6a + b = 7 \quad \text{(Equation 2)}$

Step 2: Solve the system of equations

Now, subtract Equation 1 from Equation 2: $(6a + b) - (3a + b) = 7 - 6$ $3a = 1 \implies a = \frac{1}{3}$

Substitute $a = \frac{1}{3}$ into Equation 1: $3\left(\frac{1}{3}\right) + b = 6 \implies 1 + b = 6 \implies b = 5$

So, we now have:

• $a = \frac{1}{3}$
• $b = 5$
• $c = -8$

Step 3: Find $f(-3)$

Now that we know $a$, $b$, and $c$, we can find the value of the function when $x = -3$: $f(-3) = \frac{1}{3}(-3)^2 + 5(-3) - 8$ $f(-3) = \frac{1}{3}(9) - 15 - 8 = 3 - 15 - 8 = -20$

Thus, the value of the function when $x = -3$ is $\boxed{-20}$.

Would you like further details or explanations? Here are five related questions:

1. How do you solve quadratic systems with three points?
2. Can you explain how to eliminate variables in a system of equations?
3. How does the quadratic formula relate to finding the coefficients of a parabola?
4. What are other methods for solving systems of equations?
5. Can the vertex form of a quadratic equation help in solving this?

Tip: Always start by solving for constants like $c$ first when the equation allows for an easy substitution.