Question

Find the vertex of this parabola: y=-4x^2+8x-12

Answer 1

(1,-8) is the vertex

Answer 2

The vertex of the parabola is the highest or lowest point of the graph. 

y=-4x^2+8x-12 = -4 (x^2 -2x +3) 

Lets work with this now: x^2 -2x +3

x^2 -2x +3 -> what is the closeset perfect square? 
x^2 -2x +1 = (x-1)^2
So
x^2 -2x +3 = (x-1)^2 +2

Replacing to original 
y=-4x^2+8x-12 = -4 (x^2 -2x +3) = -4 ((x-1)^2 +2) = -4 (x-1)^2 - 8

The min or max point is where the squared part = 0
So when x=1 , y= -4*0-8=-8

This will be the max of the parabola as there is - for the highest factor (-4x^2)

The max: x=1, y= -8