Prove that the extreme point of a function is at x = n? I'm attempting to learn Calculus 1 using a book that's about as clear as mud. I wish I had more time to study and understand calculus from a better and less-condensed source. 
Until then, I have been trying hard to grasp the numerous concepts flying at me one after the other in rapid succession and attempt to apply whatever methods (that haven't already been pushed out of my brain) to solve problems like the one that follows: 
"Using calculus, show that--if the function f(x) = ax^2 + bx + c has two real roots--its extreme value, maximum or minimum, occurs at x = n, where n is the midpoint between those roots."
...I apologize in advance if my logic is scatterbrained. 
Okay... well, this is a polynomial of degree 2, and it forms a parabola. It's continuous and defined everywhere, and there's no values you could plug in to make it undefined. There's no interval given either, so I think that based on this information alone, that at the point x = n, f '(x) = 0. This would be vertex of the parabola, and that's where its extreme point must be.
I'm pretty darn sure f '(x) would be
$${f(x) = ax^2 + bx + c}$$
$$f'(x) = 2ax + b$$
My first guess is to set the derivative = to 0.
$$2ax + b = 0$$
...But, I don't know if any of that matters. I mean, does the answer simply lie in proving the equation by using a theorem? If so, Rolle's Theorem perhaps? It seems to fit the bill, I think? 
Please help, I'm trying to understand this stuff. I will be appreciative of anything anyone can offer.
Many thanks,
-Jon
 A: From the derivative you can get that the extreme happens at $x=-b/2a$. 
To see that this is the midpoint of the two roots, I see two ways: either you find the two roots and take the average:
$$
\frac12\,\left( \frac{-b+\sqrt{b^2-4ax}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\right)=-\frac b{2a}.
$$
Or, but in the end it's not so different from the above, you write 
$$
f(x)=a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}\right]
$$
and you notice that $f$ is symmetric around the line $x=-b/2a$. 
Edit: a third way is to notice that $f(x)=a(x-r_1)(x-r_2)$. Expanding this we get $f(x)=ax^2-a(r_1+r_2)+ar_1r_2$. Comparing coefficients we get $$b=-a(r_1+r_2),$$, so 
$$
\frac{r_1+r_2}2=-\frac b{2a}.
$$
A: 
My first guess is to set the derivative = to 0.
  $$2ax + b = 0$$

That gives you the abscissa of the extreme value as $x = -\cfrac{b}{2a}\,$.
Now remember Vieta's relations which say that the sum of the roots of the polynomial is $-\cfrac{b}{a}\,$, then the midpoint between them is half the sum, that is $-\cfrac{b}{2a}\,$.
