This general equation you gave is a parabola that opens either up or down (U or ^ shaped, in your words) depending on the signs of $c$: if $c > 0$ it opens up, if $c < 0$ it opens down. You can get the formulas for the right/left ones (C and >) by switching $x$ and $y$, and again whether it opens right or left depending on whether $c$ is positive or negative, respectively. There many other sorts of parabolas too, though--for example, you might have a parabola at a 45-degree angle to the axes.
If I understood it correctly, the question is about the placements of separators, when the decimal expansions of elements of the Fibonacci sequence are printed on a terminal following the rule that a comma is the only separator.
The observed parabolas will appear whenever the terminal has a fixed line width $M$. A short explanation is that the length of the decimal expansion $d_n$ of $F_n$ grows linearly as a function of $n$. We get something like a parabola near the region, where $d_n\approx M$. Shortly before that point the Fibonacci numbers take slightly less than one row to print, and the commas acceleratingly drift to the right as we scroll up. Shortly after that point the Fibonacci numbers take a bit more than one row to print, and the separators begin an accelerating drift to the right as we scroll down.
Then the TL;DR; version:
This follows from Binet's formula telling us that
$$
F_n\approx\frac1{\sqrt5}\left(\frac{1+\sqrt5}2\right)^n.
$$
The approximation is very good in the sense that $F_n$ is the closest integer to the r.h.s., and the difference tends to zero quite rapidly.
Consequently $d_n$ is well approximated by
$$
d_n=1+\lfloor\log_{10}F_n\rfloor\approx 1+n\log_{10}\left(\frac{1+\sqrt5}2\right)-\log_{10}\sqrt5\approx 0.209 n+0.651.
$$
The comma following $F_n$ appears on the column $k_n, 0\le k_n<M$, where $k_n$ is determined by the congruence
$$
k_n\equiv\sum_{i=1}^n (d_i+1)\pmod M.
$$
We see that
$$
k_{n+1}-k_n\equiv d_{n+1}\pmod M.
$$
So when we are in a range, where $d_n$ is close to $M$ (the exact meaning of "close" depends on $M$), then the sum $d_n+k_n$ overflows (= exceeds $M$), and we get as the difference
$$
k_{n+1}-k_n=0.209(n-N),
$$
where $N$ is a constant. In the pictures $N$ tells us the row where the peak of the parabola lies. It is unlikely to be exactly an integer, but the small difference will not affect the conclusion below.
Namely, here the r.h.s. here is the derivative of a quadratic polynomial of the form
$$p(x)=0.1045 x^2+ax+b.$$
This explains the emergence of the parabolas. In case of a quadratic the difference, $p(n+1)-p(n)$, differs from the derivative, $p'(n)$, by a constant amount, so using one instead of the other does not change the general conclusion.
As a (slightly inaccurate) test for the above calculations I claim that the peak of the parabola will appear approximately at the comma following $F_n$, where
$$
n\approx\frac{M}{0.208988}.
$$
So if $M=80$, we get that the peak occurs after $F_{383\pm\epsilon}$.
The constant $0.208988$ is approximately $1/5$. This shows in the figures as follows. You see that the separators stay on the same column for approximately five rows, below that they drift one position to the right for the next approximately five rows. Drift two positions at a time for the next five et cetera. Scrolling up, the reverse happens.
Best Answer
It would be worth your while to learn another standard form of the equation of a parabola, and you can complete the square, given $y = ax^2 + bx + c$, to obtain this form:
$$4p(y - k) = (x-h)^2$$
The vertex of the parabola is given by $(h, k)$. $$h = \frac{-b}{2a};\quad k = \frac{4ac - b^2}{4a}$$
$$4p = \frac 1a$$