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

##### 1 Answer

Feb 12, 2016

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]}