Question

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

Answer 1

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]}