# How do you find the slant asymptote of  (x^3-1)/(x^2-6)?

Jul 4, 2016

$y = x$

#### Explanation:

$\frac{{x}^{3} - 1}{{x}^{2} - 6}$

Making ${x}^{3} - 1 = \left({x}^{2} - 6\right) \left(a x + b\right) + c x + d$

then

$\frac{{x}^{3} - 1}{{x}^{2} - 6} = a x + b + \frac{c x + d}{{x}^{2} - 6}$

as we can observe

$\frac{{x}^{3} - 1}{{x}^{2} - 6} \approx a x + b$ as $\left\mid x \right\mid$ is large.

We find $a , b , c , d$ solving

{ (-1 + 6 b - d=0), (6 a - c=0), (-b=0),( 1 - a=0) :}

Those conditions are obtained equating

${x}^{3} - 1 - \left(\left({x}^{2} - 6\right) \left(a x + b\right) + c x + d\right) = 0 \forall x$

giving $\left\{a = 1 , b = 0 , c = 6 , d = - 1\right\}$

so

The asymptote is

$y = x$