# How do you find the asymptotes for y=(x^2)/(2x^2-8)?

Feb 15, 2016

vertical asymptotes at x = ± 2
horizontal asymptote at y = $\frac{1}{2}$

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve 2x^2- 8 = 0 → 2(x^2-4) = 0 → x = ± 2

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

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

hence $y = \frac{1}{2} \text{is the equation}$
here is the graph of the function as an illustration.
graph{x^2/(2x^2-8) [-10, 10, -5, 5]}