Given the function $f(x) = -x^2 + 6x$, we want to find $f(-4)$.

To do this, substitute $x = -4$ into the function $f(x)$.

$f(-4) = -(-4)^2 + 6(-4)$

Step-by-step:

1. $(-4)^2 = 16$, so the expression becomes: $f(-4) = -16 + 6(-4)$

2. $6(-4) = -24$, so the expression is now: $f(-4) = -16 - 24$

3. Simplifying: $f(-4) = -40$

Thus, $f(-4) = -40$.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the vertex of the quadratic function $f(x) = -x^2 + 6x$?
2. How do you find the roots of $f(x) = -x^2 + 6x$?
3. How would you complete the square for $f(x) = -x^2 + 6x$?
4. What is the axis of symmetry for the graph of $f(x) = -x^2 + 6x$?
5. How does the function $f(x) = -x^2 + 6x$ compare to $g(x) = x^2 - 6x$?

Tip: When dealing with quadratic functions, always remember that the vertex form can help you identify the maximum or minimum value of the function easily.