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

Answer 1

Serenity Johnson

#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.

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Answer 2

Xavier Alonso

The horizontal asymptote of ( y = \frac{x^2-x-6}{x-2} ) is ( y = x + 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.