What is the horizontal asymptote of #y = (x^2-x-6)/(x-2)#?

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Answer 1 · 2015-10-22T23:01:04

#y = (x^2-x-6)/(x-2)#

has a vertical asymptote #x=2#

and an oblique asymptote #y = x+1#

It has no horizontal asymptote.

#y = (x^2-x-6)/(x-2)#

#= (x^2-2x+x-2-4)/(x-2)#

#=(x(x-2)+(x-2)-4)/(x-2)#

#=x+1-4/(x-2)#

As #x->+-oo#, #4/(x-2) -> 0#

So #y = x+1# is an oblique asymptote.

When #x = 2#, the denominator #(x-2)# is zero but the numerator #(x^2-x-6)# is non-zero.

So #x = 2# is a vertical asymptote.