I have this curve 
And I know that the first point is
$$A(0,5)$$
and the last point is
$$C(1650,9.5)$$
The point almost at the center where the curve changes (if you look close, you can see a green dot) is
$$B(1000,8.5)$$
There is a way to derive a generic f(x) so that I can plug in a value of x and find f(x)?
We know four properties: $f(0)=5$, $f(1000)=8.5$, $f(1650)=9.5$ and $f'(1000)=0$. To fit a polynomial with these constraints we need a degree of at least three:
$$f(x) = ax^3+bx^2+cx+d$$ $$f'(x) = 3ax^2+2bx+c$$
Plugging in what we know gives the following equation system.
$$\begin{cases} a+b+c+d=5 \\ 10^9a+10^6b+10^3c+d=8.5 \\ 1650^3a+1650^2b+1650c+d=9.5 \\ 3000000a+2000b+c=0 \\ \end{cases}$$ Solving this and plotting the resulting $f(x)$, gives the following curve. Does this look good enough?
The solution:
$$\begin{cases} a=3.562082728528 \cdot 10^{-9} \\ b=-1.063473805380 \cdot 10^{-5} \\ c=0.010583227922\\ d=4.989427403254\\ \end{cases}$$
