Answer:
Horizontal asymptote of the graph of the function f(x) = 25/1+4x is y=0
Step-by-step 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
