# How do you find the vertical, horizontal or slant asymptotes for y=(x^2)/(2x^2-8)?

Feb 26, 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 denominator equal zero.

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

equations of vertical asymptotes are x = ± 2

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

If the degree of the numerator and denominator are equal then the equation can be found by taking the ratio of leading coefficients.
Here they are both of degree 2.

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

Here is the graph of the function.
graph{x^2/(2x^2-8) [-10, 10, -5, 5]}