To me, these curves exhibit two linear regimes with a smooth transition. Denoting by $a_1x+b_1$ the first (left) linear form and by $a_2x+b_2$ the second (right) linear form and by $t$ the point, where the two linear forms meet, one could mimick such a behavior by the function
$$ f(x) = \frac{a_1x+b_1}{1+c^{x-t}} + \frac{a_2x+b_2}{1+c^{t-x}}, $$
where the exponential numerators serve as smooth step functions. The larger $c$, the sharper the transition. Here is a demonstration of how such a function looks like. For the dark blue curve, I heuristically get$a_1 \approx0.2, b_1\approx100,a_2 \approx100,b_2\approx-8.9\cdot10^4, t\approx900,c\approx1.7$.
In any case, for a rigorous fitting, it would be best to understand the origin of these curves.
3
user1010241
On
Without information on the underlying phenomenon, it is indeed difficult to make precise guesses. As Andreas says, there seems to be a transition between two linear regimes.
A possible model is
$$y=\log(e^{ax}+e^b)+cx+d.$$
It realizes a smooth transition between the slopes $c$ and $a+c$.
To me, these curves exhibit two linear regimes with a smooth transition. Denoting by $a_1x+b_1$ the first (left) linear form and by $a_2x+b_2$ the second (right) linear form and by $t$ the point, where the two linear forms meet, one could mimick such a behavior by the function $$ f(x) = \frac{a_1x+b_1}{1+c^{x-t}} + \frac{a_2x+b_2}{1+c^{t-x}}, $$ where the exponential numerators serve as smooth step functions. The larger $c$, the sharper the transition. Here is a demonstration of how such a function looks like. For the dark blue curve, I heuristically get $a_1 \approx0.2, b_1\approx100,a_2 \approx100,b_2\approx-8.9\cdot10^4, t\approx900,c\approx1.7$.
In any case, for a rigorous fitting, it would be best to understand the origin of these curves.