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What are the vertical and horizontal asymptotes of #y = (x+3)/(x^2-9)#?

Answer 1

Ethan Adkins

vertical asymptote at #x=3#

horizontal asymptote at #y=0#

hole at #x=-3#

#y = (x+3)/(x^2-9)#

First element

#y = ((x+3))/((x+3)(x-3))#

Since the factor #x+3# cancels that is a discontinuity or hole, the factor #x-3# does not cancel so it is a asymptote:

#x-3=0#

vertical asymptote at #x=3#

Let's now eliminate the factors and observe how the function changes as x grows significantly in the positive or negative:

#x -> +-oo, y ->?#

#y = cancel((x+3))/(cancel((x+3))(x-3))=1/(x-3)#

As you can see the reduced form is just #1# over some number #x#, we can ignore the #-3# because when #x# is huge it is insignificant.

We know that: #x ->+-oo, 1/x -> 0# hence, our original function has the same behavior:

#x ->+-oo, ((x+3))/((x+3)(x-3)) -> 0#

Therefore, the function has a horizontal asymptote at #y=0#

graph{y = x^2-9 [-10, 10, -5, 5]} / (x+3)

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Answer 2

Eva Abramson

Vertical asymptotes: ( x = -3 ) and ( x = 3 ) Horizontal asymptote: ( y = 0 )

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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.