# How do you find the asymptotes for y = (x^2+2x-3 )/( x^2-1)?

Mar 6, 2016

vertical asymptote x = -1
horizontal asymptote y = 1

#### Explanation:

First step is to simplify the function by factorising.

$\frac{{x}^{2} + 2 x - 3}{{x}^{2} - 1} = \frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 1\right) \left(x + 1\right)}$

and cancelling (x-1) , left with $\frac{x + 3}{x + 1}$

Vertical asymptotes occur as the denominator tends to zero. Let denominator equal zero.

solve: x+1 = 0 → x = -1 is the equation of asymptote.

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

If the degree of numerator and denominator are equal, as here they are both of degree 1. The equation can be found by taking the ratio of leading coefficients.

$y = \frac{1}{1} = 1 \Rightarrow y = 1 \text{ is the equation of asymptote }$
graph{(x^2+2x-3)/(x^2-1) [-10, 10, -5, 5]}