Question

The maximum and minimum value of
f(x) = ax^2+bx+c is found when?

Answer 1

a.) x=-b/2a
b.) x=b^2-4ac
c.) x=0
d.) the quadrartic formula
e.) all the above

Answer 2

Would it be A?

Answer 3

yes.

Answer 4

ohh okay thank you but how would it be that?

Answer 5

the vertex of ax^2+bx+c=0 is at x = -b/2a

are you allowed to use calculus if you want to prove it?

Answer 6

oh yeah that makes sense and just for future reference. how can I use calculus to prove it?

Answer 7

You find the derivative of f(x) and equate it to 0
Then solve for x

Answer 8

the maximum or minimum of f(x) occurs at f'(x) = 0
where f'(x) is the derivative of f(x)

\((ax^2+bx+c)' = 0 \\ 2ax+b = 0 \\ x = \dfrac{-b}{2a}\)

Answer 9

Ohh okay thank you!