How do you find the slant asymptote of $\frac{x^{3} - 1}{x^{2} - 6}$ $(x^3-1)/(x^2-6)$ ?

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Answer 1 · 2016-07-04T19:40:15

$y = x$ $y = x$

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

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

then

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

as we can observe

$\frac{x^{3} - 1}{x^{2} - 6} \approx a x + b$ $(x^3-1)/(x^2-6) approx ax+b$ as $| x |$ $abs(x)$ is large.

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

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

Those conditions are obtained equating

$x^3-1 -( (x^2-6)(a x+b)+c x + d)=0 forall x$

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

so

The asymptote is

$y = x$

How do you find the slant asymptote of (x^3-1)/(x^2-6)x3−1x2−6 (x^3-1)/(x^2-6)?