How do you find all the asymptotes for #(x^2 - 9)/(3x - 6)#?

1 Answer
Jul 14, 2015

Examine where the denominator is zero to find the vertical asymptote. Long divide to get a quotient linear polynomial that represents the oblique asymptote.

Explanation:

Let #f(x) = (x^2-9)/(3x-6) = ((x-3)(x+3))/(3(x-2))#

The denominator is zero when #x=2#, but the numerator is non-zero. So #f(x)# has a vertical asymptote #x=2#.

Long divide to get a quotient and remainder:

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#(x^2-9)/(3x-6) = x/3+2/3-5/(3x-6)#

Then as #x->+-oo# we have #-5/(3x-6) -> 0#

So the oblique asymptote is #y=x/3+2/3#

graph{(y-(x^2-9)/(3x-6))(y-(x/3+2/3))=0 [-10, 10, -5, 5]}