Find rhe critical poinu of y=3(x-2)^2+3.is it max or min?

Answer 1

y = 3(x-2)^2+3
y' = 6(x-2)
y" = 6

critical point is at x=2
minimum because y" > 0 there.

http://www.wolframalpha.com/input/?i=3%28x-2%29^2%2B3

But then, you knew all that from Algebra I, right? Vertex of a parabola and all that.