make an exponential curve tend to a specific value

Question

I am trying to make the following curve to exhibit an exponential decline from a given value to zero. So far, I only managed to make it go from infinite to zero.. Thus I still need to make the curve smoothly increase for smaller values, up to a preset value "top"

Here is the expression in R language:

cu=function(t){
  -(c-t)*I/(a-t)*(1+exp((b-t)*I))
}
top=2000 # maximum value for the curve
a=35 # minimum value, at which the curve should take value "top"
b=42 # value to create an inflexion point
c=45 # value to make curve zero 
I=3.4 # parameter to make the decrease in values faster

s=curve(cu,from=a,to=c,n=20)
plot(s)

Thanks in advance!

addendum: as suggested, I edited the curve to:

 cu=function(t){(-(c-a)*I/(a-t+g))*(1+exp((b-t)*I))}

but then, when using the parameters

 top=2000# 
 a=35# 
 b=42.5# 
 c=45#   
 I=3.4 # 

estimating parameter g from top

g=long/(-(c−a)*I*(1+(exp((b−a)*I))))

plotting the curve

 curve(cu,from=a,to=c,col="blue",n=10,
  ylim=c(0,long+100),
  xlim=c(34,50))

It does not stop at 2000

Answer 1

if you introduce a paraneter $g$ into the modified functional form

$$f(t)=\frac{-(c-a)I}{a-t+g}(1+e^{(b-t)I})$$

then estimating $g$ as

$$g=\frac{-(c-a)I(1+e^{(b-a)I})}{2000}$$

gives you $g=-368686801.482604$ for your specified values of $a$,$b$,$c$ and $I$.

Your curve then looks like

and you see the inflexion point at $b=42$ only when it's plotted on a log-scale, because your decay rate is rather large. I tried playing around with a multiplier on the argument of the exponential as well as varying $I$, but it didn't do much.

For a different set of parameters, the curve looks like this: