I have a curve such that as $x$ approaches infinity, $y$ approaches $2$, and as $x$ approaches negative infinity, $y$ approaches infinity. I know it contains the following approximate points:
$$(0, 70),\ (5, 40),\ (25, 11),\ (50, 2.2).$$
How can I find a function that approximates this curve? Neither logarithms nor powers under $1$ have worked for me.
On
Here's an exponential "trendline" from Excel, for the data after subtracting $2$ from the $y$ values:
It's just what @grand_chat suggested in his answer.
I wouldn't put much faith in all those decimal places in the two constants.
On
You could have a better fit using, as a model, $$y=a \exp(-b\, x^c)+2$$ which is seriously more difficult to fit compared to the case $c=1$ (in particular because of $x_1=0$).
Trying first for a few values of $c$ we have $$\left( \begin{array}{cc} c & SSQ(c) \\ 1.00 & 20.18 \\ 0.90 & 7.468 \\ 0.81 & 2.865 \\ 0.80 & 2.883 \\ 0.79 & 3.023 \\ 0.70 & 10.52 \end{array} \right)$$
Replacing $x_1=0.00001$, the nonlinear regression works fine and we get $$y=67.9542 \exp\left(-0.157416 \,x^{0.806492}\right)+2$$ corresponding to $SSQ=2.858$.
The data and predicted values are $$\left( \begin{array}{ccc} x & y & y_\text{calc}\\ 0 & 70 & 69.952 \\ 5 & 40 & 40.183 \\ 25 & 11 & 10.231 \\ 50 & 2.2 & 3.694 \end{array} \right)$$ which is not very good for the last point.
A function with exponential decay and a horizontal asymptote of y=2 will have the form $$ y= A e^{-Cx} + 2\tag1 $$ If you subtract $2$ from both sides and take the log, this gives $$ \log(y-2) =\log A - Cx\tag2 $$ Equation (2) says that there is a linear relationship between $\log(y-2)$ and $x$. So one way to fit this curve is to fit a line (either by eye, or by using a software package) to the four pairs $$ (0, \log(68)),\quad (5,\log(38)),\quad (25, \log (9)),\quad (50, \log(0.2))$$ If the line turns out to have slope $m$ and intercept $b$, then you can solve for $A$ and $C$ in (1) using the relationships $m=-C$ and $b=\log A$.