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^2-4ac)

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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