Question

find the coordinates of the vertex of the following parabola algebraically. write your answer as an (x, y) point.

y = -3x² + 6x + 12

Explanation:

Step1: Recall the formula for the x - coordinate of the vertex of a parabola in the form \(y = ax^{2}+bx + c\)

The formula for the x - coordinate of the vertex of a parabola \(y=ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). For the given parabola \(y=-3x^{2}+6x + 12\), we have \(a=-3\) and \(b = 6\).
Substitute \(a=-3\) and \(b = 6\) into the formula: \(x=-\frac{6}{2\times(-3)}\)
Simplify the right - hand side: \(x =-\frac{6}{-6}=1\)

Step2: Find the y - coordinate of the vertex

Now that we know the x - coordinate of the vertex is \(x = 1\), we substitute \(x = 1\) into the equation of the parabola \(y=-3x^{2}+6x + 12\) to find the y - coordinate.
\(y=-3\times(1)^{2}+6\times(1)+12\)
First, calculate the powers and multiplications: \(-3\times(1)^{2}=-3\times1=-3\), \(6\times(1) = 6\)
Then, add the terms together: \(y=-3 + 6+12=15\)

Answer:

\((1, 15)\)