SOLUTION: Find the value of a such that P(x)=ax^2-4x+2 has a maximum value is 16.

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Question 89204: Find the value of a such that P(x)=ax^2-4x+2 has a maximum value is 16.
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
P(x) = ax^2 - 4x + 2
vertex: (-b/2a,P(-b/2a))
vertex: (2/a,P(2/a))
P(2/a) = 16
a(2/a)^2 - 4(2/a) + 2 = 16
a(4/a^2) - 8/a = 14
4/a - 8/a = 14
-4/a = 14
-2/7 = a
~ Equation ~
P(x) = (-2/7)x^2 - 4x + 2