The coefficient of $(x-6)^2$ is negative. For a parabola, what does a negative leading coefficient tell you about whether the vertex is a maximum or a minimum?

In the form $a(x-h)^2 + k$, the vertex is at $(h,k)$. Use this to find which value of $x$ gives the extreme profit, then plug that $x$ back into $P(x)$ to get the profit itself.

In Desmos, type y = -4(x - 6)^2 + 392 {0 <= x <= 12} so the graph only shows the relevant part of the parabola.

Tap on the highest point of the parabola (the vertex) or use the function analysis tools to locate the maximum; read off the corresponding y-value, which is the maximum daily profit.

The given function is

This is a quadratic in vertex form, $a(x-h)^2 + k$. Here, $a = -4$, $h = 6$, and $k = 392$.

Because $a=-4$ is negative, the parabola opens downward, so the vertex represents a maximum value of $P(x)$.

In vertex form $a(x-h)^2 + k$, the vertex is at $(h,k)$. That means this function has its vertex at $x = h = 6$.

So the maximum profit occurs when $x = 6$ (that is, when the company produces 6 hundred units).

To find the corresponding profit, substitute $x = 6$ into $P(x)$:

Now simplify the expression for $P(6)$:

• First compute the squared term: $(6 - 6)^2 = 0^2 = 0$.
• Then multiply by $-4$: $-4 \times 0 = 0$.
• Finally add 392:

So, according to the model, the maximum possible daily profit is 392 dollars.