Math Problem Solver

\[- 3 \left(t - 0\right)^{2} + 12\]

The quadratic expression \(a (x-x_0)^2 + y_0\) is said to be in vertex form.

The quadratic expression \(a x^2 + b x + c\) can be converted to vertex form by completing the square: \[\begin{aligned}a x^2 + b x + c &= a\left(x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 \right) + c\\ &= a \left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}\\&= a(x - x_0)^2 + y_0\end{aligned}\] where \(x_0 = -\frac{b}{2 a}\) and \(y_0 = c - \frac{b^2}{4a}\).

The term vertex form comes from the fact that \((x_0, y_0)\) is the vertex of the parabola \(y = a (x-x_0)^2 + y_0\).

The quadratic expression is \(12 - 3 t^{2}\), so \(a = -3\), \(b = 0 \), and \(c = 12\).

Therefore, \(x_0 = -\frac{0}{2 \cdot -3} = 0\) and \(y_0 = 12 - \frac{0^{2}}{4 \cdot -3} = 12\).