Question 108166
You can do it two ways. 
Graphically or by analysis. 
Let's do analysis. 
As t gets large, since there is a -.125 coefficient to multiply t in the exponential, the exponential function drops quickly to zero. 
Then function then looks like
{{{B(t)=1200/(1+34e^(-0.125t))}}}
{{{B(t)=1200/(1+0)}}}
{{{B(t)=1200}}} for large positive t.
Example: At t=100, B(t)=1199.848
I'm not sure of the context of this function, that is if t is always positive, like time. 
If not, we can look at what happens when t becomes large in the negative direction. 
In that case (-.125t) becomes very large, and the exponential function becomes very large.
{{{B(t)=1200/(1+34e^(-0.125t))}}}
{{{B(t)=1200/(infinity)}}}
{{{B(t)=0}}} for large negative t. 
Example: At t=-100, B(t)=.000132
There is a horizontal asymptote at y=1200. 
And if it makes sense in your problem, there is also one at y=0.
Graphically it's easier to find the asymptotes,
{{{ graph( 300, 300, -100, 100, 0, 2000, 1200/(1+34*2.72^(-0.125*x))) }}}