The function is written in factored form with zeros at $x=-1$ and $x=7$. For a parabola, the vertex lies exactly halfway between the two zeros. What is the midpoint of $-1$ and $7$?

Once you know the $x$-coordinate of the vertex, plug that $x$ into $f(x)=a(x+1)(x-7)$. Simplify the result to get an expression involving $a$ that represents the maximum value. Then use the fact that this maximum is equal to $8$ to form an equation.

In Desmos, you can type f(x) = a(x+1)(x-7); Desmos will create a slider for a. Then, on another line, type x = 3 to show the vertical line through the vertex. Move the slider until the intersection of f(x) and x=3 has a $y$-value of $8$. The corresponding slider value of a is the solution.

On a new line, type the fraction 8/(-16). Desmos will simplify this expression; use that simplified result as the value of $a$ that satisfies the equation $-16a = 8$.

The function $f(x)=a(x+1)(x-7)$ is a quadratic.

• If $a>0$, the parabola opens up and has a minimum, not a maximum.
• If $a<0$, the parabola opens down and has a maximum.

Since the problem says the maximum value of $f$ is $8$, we already know $a$ must be negative, but we still need its exact value.

The function is given in factored form with zeros at $x=-1$ and $x=7$.

For a quadratic in factored form $k(x-r_1)(x-r_2)$, the axis of symmetry (and the vertex's $x$-coordinate) is halfway between $r_1$ and $r_2$:

Here $r_1=-1$ and $r_2=7$, so

So the maximum value of $f$ occurs at $x=3$.

Now plug $x=3$ into $f(x)$:

Compute inside the parentheses first:

• $3+1 = 4$
• $3-7 = -4$

So

Because $x=3$ is the vertex, $f(3)$ is the maximum value of the function.

We are told the maximum value of $f$ is $8$, and we just found that the maximum is $f(3) = -16a$.

So set up the equation:

Solve for $a$ by dividing both sides by $-16$:

So the value of $a$ is $-\dfrac{1}{2}$, which corresponds to choice A.