How do you find vertical, horizontal and oblique asymptotes for #(x^2-x-8)/ (x+1)#?

1 Answer
Feb 11, 2017

Answer:

vertical asymptote at x = - 1
oblique asymptote y = x - 2

Explanation:

Let #f(x)=(x^2-x-8)/(x+1)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve #x+1=0rArrx=-1" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2-x/x^2-8/x^2)/(x/x^2+1/x^2)=(1-1/x-8/x^2)/(1/x+1/x^2)#

as #xto+-oo,f(x)to(1-0+0)/(0+0)#

This is undefined, hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote. Expressing the numerator in terms of factors of the divisor.

#f(x)=(color(red)(x)(x+1)+(color(blue)(-x)-x)-8)/(x+1)#

#=(color(red)(x)(x+1)color(red)(-2)(x+1)color(blue)(+2)-8)/(x+1)#

#=x-2-6/(x+1)#

as #xto+-oo,f(x)tox-2#

#rArry=x-2" is the asymptote"#
graph{(x^2-x-8)/(x+1) [-10, 10, -5, 5]}