# calculus

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.

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1. 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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posted by Damon

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