Given an arbitrary quadratic polynomial $ax^2 + bx + c$, how do I find the path the vertex of the parabola travels upon when b is varied? https://www.desmos.com/calculator/ysuklqqwfe
I've made a desmos to illustrate the problem. If you press play on b you'll see what I mean.
I'm pretty sure the arc is parabolic because the y coordinate varies quadratically with respect to x. But I'm not sure how to go about proving it, or to find the equation of the path.
Any insights?
 A: The vertex of the parabola is$$\left(-\frac b{2a},-\frac{b^2}{4a}+c\right).$$Therefore, the set of all vertices is $\left\{(x,c-ax^2)\,\middle|\,x\in\Bbb R\right\}$, which is indeed a parabola.
A: Let $y = ax^2 + bx + c$.  For the time being, suppose $a$, $b$, and $c$ are fixed, with $a \ne 0$.
The vertex of the parabola is located where the parabola reaches an extremum, either a minimum (when $a > 0$) or a maximum (when $a < 0$).  To see where this occurs, we complete the square:
$$\begin{align}
y &= a\left(x^2 + \frac{b}{a} x\right) + c \\
&= a \left(x^2 + 2 \cdot \frac{b}{2a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}.
\end{align}
$$
As a function of $x$, keeping in mind $a, b, c$ are fixed, the only term that changes is the first; moreover, because the square of a real number is never negative, the minimum value that $(x + \frac{b}{2a})^2$ can attain is $0$, when $x = -\frac{b}{2a}$.  And as stated previously, this choice leads to a minimum when $a > 0$ and a maximum when $a < 0$.  Therefore, this is the $x$-value for the vertex, and the $y$-value is the second term, $-\frac{b^2-4ac}{4a}$; thus the vertex has coordinates
$$(x,y)_{\text{vertex}} = \left(-\frac{b}{2a}, -\frac{b^2-4ac}{4a}\right).$$
Now for a fixed $a$ and $c$, where $b$ is allowed to vary, what is the locus of the vertex--what shape does the vertices of the resulting family of parabolas trace out?  This can be realized by expressing the $y$-coordinate of the vertex in terms of the $x$-coordinate, in turn as a function of $b$.  That is to say, if $x = -b/(2a)$, then $$y = -\frac{b^2 - 4ac}{4a} = -a \left(-\frac{b}{2a}\right)^2 + c = -ax^2 + c,$$ hence the locus of the vertex is the equation $$y = -ax^2 + c,$$ also a parabola, but with the linear term equal to $0$, and facing the opposite direction to the original family of parabolas. This of course should be obvious in retrospect.
