The function `p(t)=250/(1+1.5e^((-1)/5t))` is a logistics function.

A logistics function is of the form `y=c/(1+ae^(-r(x-h)))+k` . In this case both h and k are zero, so we have `y=c/(1+ae^(-rt))` .

A logistics function in this form has two horizontal asymptotes: y=0 and y=c. The function increases on its entire domain. The y-intercept is at `(0,c/(1+a))` . Finally, the point of maximum growth is at the point `((ln a)/r,c/2)` .

For this problem a=1.5,r=`1/5` , and c=250.

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**So the point of maximum growth is at `((ln 1.5)/(1/5),250/2)` or approximately (2.03,125).**

**So the population is increasing most rapidly in year 2.**

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On the graph, you can see the functions rate of increase slowing after t=2:

Note that the y-intercept is 100, corresponding to the introduction of 100 deer at time t=0. The population is increasing the fastest when there are 125 deer in the park.

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