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

Jun 7, 2016

$y = x + 2$

Explanation:

In a polynomial fraction $f \left(x\right) = \frac{{p}_{n} \left(x\right)}{{p}_{m} \left(x\right)}$ we have:

1) vertical asymptotes for ${x}_{v}$ such that ${p}_{m} \left({x}_{v}\right) = 0$
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 $\left\mid x \right\mid$

In the present case we have

$\frac{{p}_{3} \left(x\right)}{{p}_{2} \left(x\right)} = \frac{2 {x}^{3} + {x}^{2}}{2 {x}^{2} - 3 x + 3}$
${p}_{3} \left(x\right) = {p}_{2} \left(x\right) \left(a x + b\right) + {r}_{1} \left(x\right)$
${r}_{1} \left(x\right) = c x + d$
$2 {x}^{3} + {x}^{2} = \left(2 {x}^{2} - 3 x + 3\right) \left(a x + b\right) + c x + d$

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 $\left\{a = 1 , b = 2 , c = 3 , d = - 6\right\}$
substituting in $y = a x + b$

$y = x + 2$

Note that

$\frac{{p}_{3} \left(x\right)}{{p}_{2} \left(x\right)} = \left(a x + b\right) + \frac{{r}_{1} \left(x\right)}{{p}_{2} \left(x\right)}$

and as $\left\mid x \right\mid$ increases $\frac{{r}_{1} \left(x\right)}{{p}_{2} \left(x\right)} \to 0$ 