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

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Answer 1 · 2016-02-13T16:40:08

Oblique asymptotes y= -3x +3

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

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 $x= (-6+- sqrt(36-12))/6$

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

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