How do you find the vertical, horizontal or slant asymptotes for #(x² + x - 2) /( x² + 4x + 3)#?

1 Answer
Apr 18, 2016

Answer:

vertical asymptotes x = -3 , x = -1
horizontal asymptote y = 1

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : # x^2 + 4x + 3 = 0 → (x+3)(x+1) = 0 #

#rArr x = -3 , x = -1" are the asymptotes "#

Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 #

If the degree of the numerator and denominator are equal , as in this case , then the equation is the ratio of leading coefficients, which are the coefficients of the #x^2" terms "#

#rArr y = 1/1 = 1 " is the asymptote " #

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x^2+x-2)/(x^2+4x+3) [-10, 10, -5, 5]}