Proving that if $f(x)=ax^2+bx+c$ is an even function, then $b=0$ I have a question that goes:

Prove that if $f(x)=ax^2+bx+c$ is an even function, then $b=0$.

I know the function is even because this a parabola, and it is symmetric across the y-axis.
I don't know how to prove such claim, and what b=0 means.
Can I get a clue/idea on how to solve this?
Thanks.
(I know this is a low-level question, I'm just trying to learn and get better.)
 A: We say that a function $f$ is even if for all $x$ (in the domain of $f$) we have that $f(x)=f(-x)$.
For your proof, the key is the definition. Suposse that $f(x)=ax^{2}+bx+c$ is even. Then, by definition, $f(x)=f(-x)$, i.e. $$f(x)=ax^{2}+bx+c=a(-x)^{2}+b(-x)+c=f(-x)$$Therefor $ax^{2}+bx+c=ax^{2}-bx+c$. Cancel similar terms and we obtain that $bx=-bx$. We can assume that $x\neq 0$ (why?) and therefore $b=-b$. The only number that is equal to it's negative is $0$. Thus $b=0$.
A: If the function $ f $ is even, we will have in particular
$$f(-1)=f(1)$$
which gives
$$a-b+c=a+b+c$$
and $$b=0$$
Other approach:
$$f \text{ even and differentiable at }\Bbb R \implies $$
$$f' \text{ is odd }\implies $$
$$f'(0)=-f'(0)\implies$$
$$f'(0)=0\implies$$
$$b=0 \text{ because } f'(x)=2ax+b$$
A: Not all parabolas are even. Only parabolas centered around the y-axis is even (because it's symmetrical). Parabolas centered around any other non-vertical axis are not even.
Written in vertex form, $ax^2+bx+c$ is $a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$ (expand it out and see). The x-coordinate of the vertex must be $0$ for the parabola to be centered around the y-axis and so by the property of the vertex form, the $\frac{b}{2a}$ term must be $0$. Therefore, $b=0$ as is required.
A: 
I don't know how to prove such claim, and what b=0 means.

$b=0$ means the in the function $f(x) =ax^2 + bx + c$ then the number $b$ represent is $0$.  That is to say.  If $f(x) = ax^2 + bx + c$ is an even function then $f(x) = ax^2 +0x + c = ax^2 + c$.
You have a good idea as to how to prove it.
The function $f(x) = ax^2 + bx + c = a(x + \frac b{2})^2 + (c-\frac {b^2}4)$ is symmetric on the vertical axis $x =-\frac b{2}$.  So if it is even then it is symmetric over the $y$-axis. And that can only happen if the line $x = -\frac b2$ is the $y$-axis.  As the $y$ axis is the line $x = 0$, then the line $x = -\frac b2$ con only be the same thing as the $y$-axis, if $-\frac b2 = 0$. Or in other words if $b=0$.
Of course, this assumes you have already proven that $f(x) = ax^2 + bx +c$ is symmetric across the line $x =-\frac b2$.
......
Alternatively we can literally go but to definitions.
$f(x)$ is even if $f(x) = f(-x)$ for all $x$.
So $a(-x)^2 + b(-x) + c = ax^2 + bx + c$ for all $x$ so
$ax^2 -bx + c = ax^2 + bx + c$ for all $x$ so
$-bx = bx$ for all $x$ so
$2bx = 0$ for all $x$.
$bx = 0$ for all $x$.
The only way that is possible is if $b=0$.
