find the horizontal asymptote of f(x)= 8x/2x^2+1
as x gets huge, all but the highest powers don't matter, so you have
f(x) -> 8x/2x^2 = 4/x -> 0
whenever the denominator is a higher power than the numerator, the asymptote is y=0.
To find the horizontal asymptote of the function f(x) = 8x / (2x^2 + 1), we need to analyze the behavior of the function as x approaches positive or negative infinity.
First, let's look at the degrees of the numerator and the denominator. The degree of the numerator is 1 (due to x), and the degree of the denominator is 2 (due to x^2).
Since the degree of the denominator is greater than the degree of the numerator, we can determine the presence of a horizontal asymptote by dividing the coefficients of the highest degree terms. In this case, that means dividing 8 by 2.
So, the horizontal asymptote is given by the ratio of these coefficients, which is 8/2 = 4.
Therefore, the horizontal asymptote of the function f(x) = 8x / (2x^2 + 1) is y = 4.