Question

How do you find the oblique (or slant) asymptote for`( 2x^3 + x^2) / (2x^2 - 3x + 3)`?

Answer 1

`y = x +2`

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

equating coefficients

`{ (-3 b - d=0), (-3 a + 3 b - c=0), (1 + 3 a - 2 b=0), (2 - 2 a=0) :}`

solving for `a,b,c,d` we have `{a = 1, b = 2, c = 3, d = -6}`
substituting in `y = a x + b`

`y = x +2`

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`