How do you find vertical, horizontal and oblique asymptotes for $(x^3-8)/(x^2-5x+6)$ ?

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Answer 1 · 2018-06-13T21:43:39

the oblique asymptote is $y=x+5$ and the horizontal asymptote is $x=3$

$(x^3-8)/(x^2-5x+6)=(cancel(x-2)(x^2+2x+4))/((x-3)cancel(x-2))$
= $(x^2+2x+4)/(x-3)=((x-3)(x+5)+19)/(x-3)=x+5+19/(x-3)$

Therefore, the oblique asymptote is $y=x+5$ and the horizontal asymptote is $x=3$

The horizontal asymptote is found by letting the denominator equal to 0.

The oblique asymptote is found by long division

How do you find vertical, horizontal and oblique asymptotes for (x^3-8)/(x^2-5x+6)(x^3-8)/(x^2-5x+6)?