How do you find the horizontal asymptote for #(x^2+1)/(2x^2-3x-2)#?

1 Answer
Feb 12, 2016

Answer:

horizontal asymptote at y = #1/2 #

Explanation:

horizontal asymptotes occur as #lim_(x→±∞ )f(x) → 0#

If the degree of the numerator and denominator are equal , as they are in this question, both degree 2 , then the equation can be found by taking the ratio of leading coefficients.

# rArr y = 1/2 #

here is the graph of the function as an illustration.
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}