Question 53314
find the vertex of:  f(x)=3x^2-6x+8
Rewrite as: 
3x^2-6x+? = y-8
Rewrite as:
3(x^2-2x+?=y-8
Rewrite as :
(x^2-2x+? =(1/3)(y-8)
Complete the square:
x^2-2x+1 = (1/3)(y-8)+1
(x-1)^2=(1/3)(y-8+3)
(x-1)^2=(1/3)(y-5)
Vertex is at (1,5)
Cheers,
Stan H.