How do you find vertical, horizontal and oblique asymptotes for #( x²-x-12 )/(x+5)#?

1 Answer
Oct 21, 2016

Answer:

The vertical asymptote is #x=-5#
The oblique asymptote is #y=x-6#

Explanation:

The domain of the function is #RR-(5)#
as we cannot divide by #0#
So the vertical asymptote is #x=-5#
The degree of the polynomial of the numerator #># degree of the denominator, so we have an oblique asymptote
doing a long division give

#(x^2-x-12)/(x+5)=x-6+(18)/(x+5)#

So #y=x-6# is an oblique asymptote

Verification

#x-6+(18)/(x+5)=((x-6)(x+5)+18)/(x+5)=(x^2-x-30+18)/(x+5)#

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