Asked by Ema
Find rhe critical poinu of y=3(x-2)^2+3.is it max or min?
Answers
Answered by
Steve
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.
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.
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