How can we assume that the all parabolas in the form 'ax^2 + bx +c' are symmetrical? So I was reading an answer to a question pertaining to the derivation of the line of symmetry. It reads as follows:

The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down. So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry. Factoring $y=x(ax+b)$, we see that the x-intercepts of this parabola occur at $x=0$ and $x=\frac{-b}a$, and hence the axis of symmetry lies halfway between, at $x=\frac{−b}{2a}$.

This proof assumes that the vertex of a parabola is exactly between the two roots of the parabola (i.e. the two $x$ values when $y = 0$) so therefore a line of symmetry exists for all parabolas in the form $ax^2 + bx + c$. How can we fully ascertain this assumption; I can envisage a curved line where the vertex is skewed towards one root of the curve. Why is this not possibly for a parabola in the form $ax^2 + bx + c$?
 A: Well, consider what symmetry means; suppose we have a function
$$f(x)=ax^2+bx+c.$$
Then, if it is symmetrical about some point $v$, we should have:
$$f(v+x)=f(v-x)$$
for any other $x$ - that is, the function takes on the same value a fixed distance to the right of $v$ as it does to the left of $v$. If we want to prove all parabolas have symmetry, we just need to show that such a $v$ exists.
In particular, we can expand the equation $f(v+x)=f(v-x)$ as requiring the following for all $x$:
$$a(v+x)^2 + b(v+x) + c = a(v-x)^2 + b(v-x) + c$$
Expanding each side above and collecting monomials in $x$ yields:
$$ax^2 + (b+2va)x + av^2 + bv + c = ax^2 + (-b-2va)x + av^2 + bv + c$$
and then canceling like terms yields:
$$(b+2va)x = (-b-2va)x.$$
This last equation is equivalent to $f(v+x)=f(v-x)$ and it clearly holds for all $x$ if $b+2va=0$ - that is, if $v=\frac{-b}{2a}$. This suffices to prove that all parabolas are symmetric about the line $x=\frac{-b}{2a}$.
A: Given the function $f(x)=ax^2+bx+c$, consider the transformation $$g(x)=f(x-\frac{b}{2a})$$ 
Now consider that, when simplified, $g(x)=ax^2+c$, for which $g(-x)=g(x)  \forall x$, which means that $g(x)$ is an even function with line of symmetry which is the $y$ axis.
And since $f(x)$ is a translation of $g(x)$, we can see that $f(x)$ has a line of symmetry $x=-\frac{b}{2a}$
