Identify the Axis of Symmetry for f(x) = -3x^2 + 12

To solve this problem, we'll follow these steps:

• Step 1: Identify the given coefficients from $f(x) = ax^2 + bx + c$: $a = -3$, $b = 0$, and $c = 12$.
• Step 2: Apply the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex.
• Step 3: Plug the x-coordinate back into the function to find the y-coordinate.

Now, let's work through each step:
Step 1: The function is given as $f(x) = -3x^2 + 12$. We have $a = -3$, $b = 0$, and $c = 12$.
Step 2: Using the formula $x = -\frac{b}{2a}$, substitute $b = 0$ and $a = -3$:
$x = -\frac{0}{2 \times -3} = 0$
Step 3: Substitute $x = 0$ back into the quadratic function to find the y-coordinate:
$f(0) = -3(0)^2 + 12 = 12$
So, the vertex, or symmetry point, of the function is $(0, 12)$.

Therefore, the solution to the problem is $(0, 12)$.