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.