How do you find the vertical, horizontal and slant asymptotes of: #g(x) = (x²-x-12) /( x+5)#?

1 Answer
Nov 28, 2016

Answer:

vertical asymptote at x = - 5
slant asymptote is y = x - 6

Explanation:

The denominator of g(x) cannot be zero as this would make g(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+5=0rArrx=-5" is the asymptote"#

Horizontal asymptotes occur as

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

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

#g(x)=(x^2/x^2-x/x^2-12/x^2)/(x/x^2+5/x^2)=(1-1/x-12/x^2)/(1/x+5/x^2)#

as #xto+-oo,g(x)to(1-0-0)/(0+0)to1/0#

This is undefined hence there are no horizontal asymptotes.

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is a slant asymptote.

#"Using "color(blue)"polynomial division"#

#g(x)=x-6+18/(x+5)#

as #xto+-oo,g(x)tox-6+0#

#rArry=x-6" is the asymptote"#
graph{(x^2-x-12)/(x+5) [-10, 10, -5, 5]}