Answer: A

The zeros of $f$ are $4$ and $-2$, so the vertex is halfway between them, at $x = \frac{4 + (-2)}{2} = 1$. Because $f$ has a maximum, that maximum occurs at the vertex: $f(1) = a(1 - 4)(1 + 2) = -9a$. Setting $-9a = 27$ gives $a = -3$. A negative $a$ makes the parabola open downward, which agrees with $f$ having a maximum.

Choice D has the right size but the wrong sign. With $a = 3$, the parabola opens upward and has a minimum of $-27$, not a maximum of $27$.

Why the other choices are wrong

• B. This divides -9 by 27 instead of dividing 27 by -9. From -9a = 27, a = -3.
• C. A positive a makes the parabola open upward, so f would have a minimum, not a maximum.
• D. With a = 3, the parabola opens upward and f(1) = -27 is a minimum, not a maximum of 27.