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

1 Answer
Gió
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=-b/(2a)#
using your equation in the form #y=ax^2+bx+c# where:
#a=-1#
#b=4#
#c=12# you get:
#x_v=-4/(-2)=2#
Substituting back this value into your equation you get:
#y_v=-4+8+12=16#

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