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

Feb 13, 2016

Oblique asymptotes y= -3x +3

Vertical asymptotes $x = \frac{- 3 \pm \sqrt{3}}{3}$

#### Explanation:

On division the given expression appears like this

(-3x +3)+ $\frac{16 x + 3}{3 {x}^{2} + 6 x + 2}$

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

Vertical asymptotes are given by $3 {x}^{2} + 6 x + 2 = 0$, that is $x = \frac{- 6 \pm \sqrt{36 - 12}}{6}$

$x = \frac{- 3 \pm \sqrt{3}}{3}$