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

Mathematics

1 answer:

const2013 [10]4 years ago

5 0

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

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