Question

In Exercise,find the horizontal asymptote of the graph of the function.
f(x) = 25/1+4x

Answer 1

Horizontal asymptote of the graph of the function f(x) = 25/1+4x is y=0

Explanation:

I attached the graph of the function. Graphically it can be seen that the horizontal asymptote of the graph of the function is y=0. There is also a vertical asymptote at x=-1/4

When denominator's degree (1) is higher than nominator's degree (0) then the horizontal asymptote is y=0