A quadratic function ‒x2 + bx + c attains its maximum value at x = 4. If f(‒8) = 0, then find f(10).

1. Break Down the Problem

We have a quadratic function of the form: $f(x) = -x^2 + bx + c$

It is given that the function attains its maximum value at $x = 4$. For a quadratic function in the form $ax^2 + bx + c$, the vertex (maximum value for $a < 0$) is given by the formula: $x = -\frac{b}{2a}$ Since $a = -1$ in this case, we can rewrite this as: $4 = -\frac{b}{2(-1)} = \frac{b}{2}$

We also know that $f(-8) = 0$, which gives another equation to work with.

2. Relevant Concepts

From the vertex formula, we can find the value of $b$: $4 = \frac{b}{2} \implies b = 8$

Now substituting into $f(x)$, we get: $f(x) = -x^2 + 8x + c$

Next, we will use the condition $f(-8) = 0$ to find the value of $c$: $f(-8) = -(-8)^2 + 8(-8) + c = 0$ $-64 - 64 + c = 0$ $c = 128$

3. Analysis and Detail

Now we have determined both $b$ and $c$:

• $b = 8$
• $c = 128$

The full function is: $f(x) = -x^2 + 8x + 128$

Next, we need to calculate $f(10)$: $f(10) = -(10)^2 + 8(10) + 128$ $= -100 + 80 + 128$ $= -100 + 208$ $= 108$

4. Verify and Summarize

All calculations have been verified following the established formulas.

Final Answer

Thus, the value of $f(10)$ is: $\boxed{108}$