How do you find the asymptotes for #{-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}#?

1 Answer
Feb 13, 2016

Answer:

Oblique asymptotes y= -3x +3

Vertical asymptotes #x=(-3+-sqrt 3)/3#

Explanation:

On division the given expression appears like this

(-3x +3)+ # (16x+3)/(3x^2 +6x +2)#

Oblique asymptote is given by y=-3x+3

Vertical asymptotes are given by #3x^2 +6x+2=0#, that is #x= (-6+- sqrt(36-12))/6#

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