Was the quadratic function derived?

Question

By quadratic function, I mean :

$$f(x)=ax^2+bx+c$$

Because the specific constant have a different role, e.g. increasing $a$ has a role of narrowing or inverting the parabola, I'm thinking if this was a derived function/equation. If so, how was this derived?

Additionally, for an even more general equation of a circle for instance:

$$x^2+y^2+Dx+Ey+F=0$$

What would be the role of each of the constants: $D$, $E$ and $F$?

And how can I go further into this?

Answer 1

As for the additional part of your question, for the following equation of circle, $$x^2+y^2+Dx+Ey+F=0$$ the co-ordinates of the center of the circle are $(-\frac{D}{2}, -\frac{E}{2})$ (the role of the constants $D$ and $E$ right there) and the radius of the circle is $\sqrt{\frac{D^2}{4} + \frac{E^2}{4}-F}$; so for fixed $D$ and $E$, $F$ is inversely proportional to the radius(size) of the circle.

EDIT:
Here is an "independent" geometric meaning of the constant $F$: square of the length of the tangent to the circle from the origin.
$1$. If the origin lies outside the circle then $2$ tangents of equal length can be drawn to the circle. Here $F$ is positive and it's square root gives the lengths of these tangents.
$2$. If the origin is inside the circle, then $F$ is negative. Its square root(and hence the length) is imaginary. Hence no tangent can be drawn to the circle from origin in this case.
$3$. If the origin is on the circle($F=0$) then, length of the tangent from it is $0$. Note that length in all these cases denote the distance from the origin to the point of tangency. Here these points coincide. Hence the length is $0$.

Answer 2

I think you're seeing this the wrong way around. It wasn't that we chose the standard form $ax^2+bx+c$ because we could change the geometry. The choice was simply the most natural way of writing a quadratic. Once we had the most simple form, we then looked at the geometric implications of altering the constants.