How do you find vertical, horizontal and oblique asymptotes for #(x + 1) / (x^2 - x -6)#?

1 Answer
Apr 5, 2016

vertical asymptotes x = -2 , x = 3
horizontal asymptote y = 0

Explanation:

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

solve: #x^2 - x -6 = 0 → (x-3)(x+2) = 0 #

#rArr x = -2 , x = 3 " are the asymptotes "#

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

When the degree of the numerator < degree of denominator the equation of the asymptote is always y = 0.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(x+1)/(x^2-x-6) [-10, 10, -5, 5]}