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

1 Answer
Mar 6, 2016

Answer:

vertical asymptote x = -1
horizontal asymptote y = 1

Explanation:

First step is to simplify the function by factorising.

#(x^2+2x -3)/(x^2 - 1) =( (x+3)(x-1))/((x-1)(x+1))#

and cancelling (x-1) , left with #(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 = 1/1 = 1 rArr y = 1 " is the equation of asymptote " #
graph{(x^2+2x-3)/(x^2-1) [-10, 10, -5, 5]}