The given function is:

$f(x) = \frac{3x^2 + 3x - 12}{3}$

First, simplify the equation by dividing every term inside the parentheses by 3:

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

Now, to write this in vertex form, we will complete the square.

1. Start with the equation: $f(x) = x^2 + x - 4$

2. Focus on the quadratic and linear terms: $x^2 + x$. To complete the square, take half of the coefficient of $x$, square it, and add/subtract it inside the equation.

The coefficient of $x$ is 1. Half of 1 is $\frac{1}{2}$, and squaring it gives $\frac{1}{4}$.

3. Add and subtract $\frac{1}{4}$ inside the equation:

$f(x) = (x^2 + x + \frac{1}{4}) - \frac{1}{4} - 4$

4. Now, rewrite the trinomial $x^2 + x + \frac{1}{4}$ as a perfect square:

$f(x) = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} - 4$

5. Combine constants $- \frac{1}{4}$ and $-4$:

$f(x) = \left(x + \frac{1}{2}\right)^2 - \frac{17}{4}$

Thus, the vertex form of the equation is:

$f(x) = \left(x + \frac{1}{2}\right)^2 - \frac{17}{4}$

Let me know if you'd like further explanation or have any questions.

Relative Questions:

1. What is the vertex of this quadratic function?
2. How do you find the axis of symmetry from the vertex form?
3. Can you solve for the x-intercepts using the vertex form?
4. What is the significance of the constant term in the vertex form?
5. How does completing the square help in solving quadratic equations?

Tip:

In vertex form, the coordinates of the vertex $(h, k)$ are easily identified as $(-h, k)$ from the expression $f(x) = a(x-h)^2 + k$.