How do you find the VERTEX of a parabola y=-x^2+4x+12?

1 Answer
Jul 18, 2015

I found:
${x}_{v} = 2$
${y}_{v} = 16$

Explanation:

You could use the derivative (setting it equal to zero to find the $x$ coordinate of the vertex) but I am not sure you know the derivative so, instead, you can find the $x$ coordinate using the relationship:
${x}_{v} = - \frac{b}{2 a}$
using your equation in the form $y = a {x}^{2} + b x + c$ where:
$a = - 1$
$b = 4$
$c = 12$ you get:
${x}_{v} = - \frac{4}{- 2} = 2$
Substituting back this value into your equation you get:
${y}_{v} = - 4 + 8 + 12 = 16$