# How do you find the asymptotes for f(x) = (5x)/(x^2-1)?

Feb 23, 2016

vertical asymptotes at x = ± 1
horizontal asymptote at y = 0

#### Explanation:

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

solve  x^2 - 1 = 0 → (x-1)(x+1) = 0 → x = ± 1

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

If the degree of the numerator is less than the degree of the denominator, as is the case here, numerator degree 1 , denominator degree 2.

Then the equation is y = 0

Here is the graph as an illustration of the asymptotes.
graph{5x/(x^2-1) [-10, 10, -5, 5]}