Question

How do you find vertical, horizontal and oblique asymptotes for `{-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}`?

Answer 1

Vertical asymptotes `x_v = {1/3 (-3 - sqrt[3]), 1/3 (-3 + sqrt[3])}`
Slant asymptote `y = 3 - 3 x`.

Explanation:

In a polynomial fraction `f(x) = (p_n(x))/(p_m(x))` we have:

`1)` vertical asymptotes for `x_v` such that `p_m(x_v)=0`. Here `x_v = {1/3 (-3 - sqrt[3]), 1/3 (-3 + sqrt[3])}`
`2)` horizontal asymptotes when `n le m`
`3)` slant asymptotes when `n = m + 1`
In the present case we dont have vertical asymptotes and `n = m+1` with `n = 3` and `m = 2`

Slant asymptotes are obtained considering `(p_n(x))/(p_{n-1}(x)) approx y = a x+b` for large values of `abs(x)`

In the present case we have

`(p_3(x))/(p_2(x)) =(-9 x^3 - 9 x^2 + 28 x + 9)/(3 x^2 + 6 x + 2)`
`p_3(x)=p_2(x)(a x+b)+r_1(x)`
`r_1(x)=c x + d`
`-9 x^3 - 9 x^2 + 28 x + 9 = (a x + b) (3 x^2 + 6 x + 2) + c x + d`

equating coefficients

`{ (9 - 2 b - d=0), (28 - 2 a - 6 b - c=0), (9 + 6 a + 3 b=0), (9 + 3 a=0) :}`

solving for `a,b,c,d` we have `{a = -3, b = 3, c = 16, d = 3}`
substituting in `y = a x + b`

`y = 3 - 3 x`

Note that

`(p_3(x))/(p_2(x))=(a x+b)+(r_1(x))/(p_2(x))`

and as `abs(x)` increases `(r_1(x))/(p_2(x))->0`

Attached a figure showing the results.