Posted by Janet on Sunday, May 8, 2011 at 11:34pm.
Quadratic function written in standard form where a, b, and c are constants such that a is not zero.
f(x)= ax^2+bx+c
Using calculus find the vertex of the parabola formed by this quadratic function. Determine under what conditions this vertex is a maximum or minumum (using Calculus techniques). Show work using derivatives to justify our conclusions.
THis is what I did so far, but can't figure out the rest.
f'(x)= 2ax + b
0=2ax+b
b = 2ax
b/2a = x
Please help and show steps so that I can understand.

calculus  Damon, Monday, May 9, 2011 at 6:19am
That is correct. The minimum ofr maximum (vertex) is at x = b/2a
You already know that from the quadratic equation. x = b/2a +/ (b/2a)sqrt (b^24ac)
Now
if x = b/2a
what is y?
y = a (b^2/4a^2) + b(b/2a) + c
= b^2/4a b^2/2a + c
= b^2/4a+c
for max or min
f" = 2 a
if a is +, that is a minimum
if a is , that is a maximum
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