PLEASE HELP!The moose population in a New England forest can be modeled by the function y = 60x/ 1 + 0.625x . What is the value of the horizontal asymptote? Describe it's meaning in the context of the problem. A) y = 96; The maximum number of moose that the forest can sustain at one time.  B) x = 96; The maximum number of moose that the forest can sustain at one time. C) x = -1.6; The minimum number of moose that the forest can sustain at one time. D) y = -1.6; The minimum number of moose that the forest can sustain at one time.

You have the function [tex]y= \dfrac{60x}{1+0.625x} [/tex]. 
Rewrite it in the following way:

[tex]y= \dfrac{60x}{1+0.625x}=\dfrac{60x}{1+ \frac{5}{8} x}=\dfrac{480x}{8+5x}=\dfrac{96\cdot 5x}{8+5x}=[/tex]
[tex]=\dfrac{96(5x+8-8)}{8+5x}=\dfrac{96(5x+8)}{8+5x}-\dfrac{96\cdot8}{8+5x}=96-\dfrac{768}{8+5x}[/tex].
This function entry allows you to determine that y=96 is the horizontal asymptote and x=-8/5 is the vertical asymptote.

From the added graph of the function you can conclude that the maximum number of moose that the forest can sustain at one time is 96.
Answer: Correct choice is A.