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

Feb 12, 2016

horizontal asymptote at y = $\frac{1}{2}$

#### 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.

$\Rightarrow y = \frac{1}{2}$

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