Question

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.

Answer 1

You have the function
`y= (60x)/(1+0.625x)`.
Rewrite it in the following way:


`y= (60x)/(1+0.625x)=(60x)/(1+ (5)/(8) x)=(480x)/(8+5x)=(96\cdot 5x)/(8+5x)=`

`=(96(5x+8-8))/(8+5x)=(96(5x+8))/(8+5x)-(96\cdot8)/(8+5x)=96-(768)/(8+5x)`.
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.