Question 47515
Determine the horizontal asymptote of the grahp of B(t) = 1200 divided by 1+34e^-0.125t


AS t goes to infinity, e^-0.125t goes to zero.

So B(t) = 1200/[1+34e^-0.125t] goes to 1200/1 = 1200

Cheers,
Stan H.