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

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Answer 1 · 2016-03-20T06:32:44

Vertical Asymptote: $x=3$
Oblique Asymptote: $y=x+5$
Horizontal Asymptote: None

To obtain the oblique asymptote, we divide by long division

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

the linear part (x+5) of the quotient will used for the oblique asymptote, that is

$y=x+5$

The following includes the oblique asymptote $y=x+5$ and the given $y=(x^3-8)/(x^2-5x+6)$

graph{(y-(x^3-8)/(x^2-5x+6))(y-x-5)=0[-80,80,-40,40]}

God bless....I hope the explanation is useful.

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