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

##### 1 Answer

Feb 26, 2016

vertical asymptotes at x = ± 2

horizontal asymptote at

#### 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.

#rArr y = 1/2 # Here is the graph of the function.

graph{x^2/(2x^2-8) [-10, 10, -5, 5]}