Is there any equation for this type of skewed parabola?

Question

I am having the following parabola looking curve (blue curve), but its not exactly symmetric. The best fit that I am getting using a quadratic equation is also shown (black curve) but it is not perfectly fitting it.

Is there any simple equation for fitting such a skewed parabolic curve?

The data is from experiment. Y-axis is flow velocity in vertical direction (in m/s), and X-axis is the depth of water (in mm). The parabola vertex can be assumed to be at 'D' as shown in the 2nd figure.

Thanks a lot.

Edit 1: Below is a pic of the experiment set-up in the question for easy understanding. The axes are interchanged in the figures.

Answer 1

As far as your diagram looks, you might try some ombination of parabolas to fit your function..

Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like $f(x)$=Parabola A when y coordinates below a certain point and Parabola B otherwise.

Now what these two (or more) parabolas are, you yourself have to find out , by brute force or something..

Answer 2

This is a fairly old question, but I wanted to share a nice solution offered in this thread. The solution they give was to use a Linex function, which looks like that:

$f(x) = \exp(ax) - ax - 1$

Which, when plotted, looks like that:

Note that I scaled the functions to compare their "skewness". Python code to reproduce the plot:

import numpy as np
from matplotlib import pyplot as plt

x = np.linspace(-2,2)
y = lambda a : np.exp(a * x) - a * x - 1
y_scaled = lambda a : (y(a) - np.min(y(a))) / (np.max(y(a)) - np.min(y(a)))

plt.plot(x, -y_scaled(1.25), x, -y_scaled(0.75), x, -y_scaled(0.25))
plt.legend(['a = 1.25', 'a = 0.75', 'a = 0.25'])
plt.show()

This, of course, doesn't at all look like a parabola (well, it is an exponential function), but did the trick for me when I tackled a similar problem: