I was looking at Cosine and I thought it was cool, I was wondering what the formula for a horizontal version of it would look like.
A graph for Something like this :
[Math] Formula for a vertical cosine graph
graphing-functions
Related Solutions
It would help to know more criteria since there are a lot of possible functions here. The hyperbola has been mentioned. Two more examples with some different behavior are
$$y = 20 - 1000\frac{\exp(-\frac{x}{8})}{x}~ (in~ green~ below)\\
y = \frac{20x^{2}+x-10}{x(x+10)}~ (in~ black~ below)$$
The hyperbola solution is in red below.
Basically $y = 20 - \frac{f(x)}{g(x)}$, up to tweaking signs of terms, will work if $g(0) = 0$ and $g$ has no other positive roots, and $\lim_{x\rightarrow +\infty} f(x) = 0$. Within reason you can get a lot of behaviors.
Edit: it appears the OP has some additional slope constraints. Basically every slope constraint requires a new coefficient. Consider the rational function $$R(x) = \frac{Ax^n + a_1x^{n-1} + \cdots - a_n}{x^n + x^{n-1} + \cdots + dx}, \forall i \in [1,n]~ (a_i \geq 0)$$ Only $a_n$ is marked as negative. Note this will match the criteria in the OP: $$\lim_{x\rightarrow +\infty}R(x) = A \\ \lim_{x\rightarrow 0^+}R(x) = -\infty$$ The derivative of $R(x)$ is relatively straightforward. Each variable we choose to be non-zero will allow us to specify a different slope at a different point; for instance, if we demand that $R'(1) = 1$ and we set $a_n=2$ then we can determine $d$ algebraically. If we instead demand only that $R'(1) \approx 1$ this gives us some more wiggle room to pick "nice" numbers. But in any case additional constraints require additional coefficients and as more conditions are added the exponent will start to rise in order to give us additional coefficients needed to fit our slope parameters.
But since we can always introduce more variables and higher exponents, we can always solve the system of equations demanded by all the required slopes.
Consider your standard logistic function
$$y = \frac{L}{1+e^{-k(x-x_0)}}$$
Filling in the data about minimum, maximum, and midpoint, we get
$$y = \frac{100}{1 + e^{-k(x-3000)}}$$
So it remains to find the correct value for "steepness" which will pass through the remaining point.
Plugging in our other point, we get
$$75 = \frac{100}{1 + e^{-k(4500-3000)}}$$
$$\frac{3}{4} = \frac{1}{1 + e^{-1500k}}$$ $$\frac{4}{3} = 1 + e^{-1500k}$$ $$e^{-1500k} = \frac{1}{3}$$ $$e^{1500k} = 3$$ $$k = \frac{\ln{3}}{1500} \approx 0.00073240819$$
Which we can confirm by plugging back in to the earlier equation.
Best Answer
A simple one would be $x=cos(y)$.