Why do the vertices of $f(x) = ax^2 + bx + c$, when fixing $a$ and $c$ but varying $b$, lie on $g(x) = -ax^2 + c$? As a bit of background context, I'm teaching a calculus course this semester, and for a learning activity, the students had to examine various functions and how certain modifications affected them.
In one case, they were examining the function $f(x) = 3x^2 + bx - 2$, for various values of $b$: what did those changes in $b$ do to the shape of the parabola, and why?
I got curious, and noticed that as we varied $b$, the parabola would move in a parabolic way with respect to the vertex. That is, if we increased $b$ when it was positive, it would move right and down; if $b$ was decreased and negative, it would move left and down. (The vertical changes were sharper for larger $b$ of course.)
So naturally, I wanted to graph the various vertices of the function. I couldn't figure out a proper way to do this in Desmos, but I managed something feasible:

*

*Create a list $L$ of test values from $-100$ to $100$ with small intervals in-between.

*These $L$ will function as our $b$ values.

*The axis of symmetry of $f$ then is at $x = -b/6$ for each $b \in L$.

*Of course, then, the vertex is at $(-b/6,f(-b/6))$ for each $b \in L$.

(You can play with a demo of it here if you wish.) Graphing a bunch of these points in black and $\color{red}{f(x)}$ in red gives this:

Curiously, the set of vertices (for variable $b$) lie on a parabola that is precisely $f$ (when $b=0$) flipped vertically; that is, the vertices lie on $g(x) = -3x^2 - 2$. (And in general, if we vary $b$ in $f(x) = ax^2 + bx + c$, then the vertices lie on $g(x) = -ax^2 + c$.)
I've been wondering about this for a few days and a student even brought it up to me, and I can't think of a reason why this should be. Maybe it's incredibly simple, but can someone perhaps enlighten me? Thanks!
 A: The coordinates of the vertex are

*

*$x=-\frac b{2a}$

*$y=-\frac{b^2-4ac}{4a}=-a\left(\frac b{2a}\right)^2+c$
therefore
$$y=-ax^2+c$$
A: Rewrite your function so you immediately see where the vertex is

*

*$ f(x)=a\left(x^2+\frac b ax+
   \frac{b^2}{(2a)^2}-\frac{b^2}{(2a)^2}\right)+c$

*$f(x)=a\left(x+\frac{b}{2a}\right)^2 +c-\frac{b^2}{4a}$
so you see how the vertex $ \left(-\frac{b}{2a};c-\frac{b^2}{4a}\right)$ changes with b and c.
A: Varying $b$ is like shifting the parabola around, always keeping the parabola through the point $(0, c)$.
Then not only the vertex traces the same parabola shape; pick any point on the parabola, when varying $b$, that point will trace the same inverted parabola shape.
A: For parabola $ y = a x^2+b x +c $
$$ x_{@Vertex}=\dfrac{-b}{2a}$$
$$ y_{extremum}= c- \dfrac{b^2}{4a}$$
Eliminate  $b$ between them and the dotted curve in demo shows locus as :
$$ y_{extremum}= c- a x_{@Vertex}^2$$
If familiarized with envelopes/ singular solutions, we would also  note that all moving parabolas pass through $(0,-2),$ as a singular point of the family, by C-discriminant method involving partial derivative and parameter $b$ elimination.
