PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!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.
Accepted Solution
A:
Answer: A & B are the same answer --> 96, maxStep-by-step explanation:Consider m is the degree of the numerator (top) and n is the degree of the denominator (bottom). Then the horizontal asymptote (H.A.) is based on the relationship between m and n:If m > n, then there is no H.A.If m = n, then y = coefficient of numerator ÷ coefficient of denominatorIf m < n, then y = 0In the given problem, m = 1 and n = 1 so the H.A. is:[tex]y=\dfrac{60}{0.625}\quad \rightarrow \quad y=96[/tex]This is the maximum number of moose that the forest can sustain at one time.