How do you find the asymptotes for $f (x) = \frac{x^{3} - 1}{x^{2} - 6}$ $f(x) =(x^3-1)/(x^2-6)$ ?

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Answer 1 · 2017-01-10T07:50:15

The vertical asymptotes are $x = - \sqrt{6}$ $x=-sqrt6$ and $x = \sqrt{6}$ $x=sqrt6$
The slant asymptote is $y = x$ $y=x$
No horizontal asymptote

Let's factorise the denominator

$x^{2} - 6 = (x - \sqrt{6}) (x + \sqrt{6})$ $x^2-6=(x-sqrt6)(x+sqrt6)$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{-sqrt6,sqrt6}$

As we cannot divide by $0$ , $x!=-sqrt6$ and $x!=sqrt6$

The vertical asymptotes are $x=-sqrt6$ and $x=sqrt6$

As the degree of the numerator is $>$ than the degree of the denominator, there is a slant asymptote.

We perform a long division

$color(white)(aaaa)$ $x^3$ $color(white)(aaaaa)$ $-1$ $color(white)(aaaaa)$ $∣$ $x^2-6$

$color(white)(aaaa)$ $x^3-6x$ $color(white)(aaaa)$ #color(white)(aaaa)∣# $x$

$color(white)(aaaaa)$ $0+6x-1$ $color(white)(aaaa)$

Therefore,

$f(x)=(x)+(6x-1)/(x^2-6)$

$lim_(x->-oo)(f(x)-x)=lim_(x->-oo)(6x-1)/(x^2-6)=lim_(x->-oo)(6x)/x^2=lim_(x->-oo)6/x=0^-$

$lim_(x->+oo)(f(x)-x)=lim_(x->+oo)(6x-1)/(x^2-6)=lim_(x->+oo)(6x)/x^2=lim_(x->+oo)6/x=0^+$

So, the slant asymptote is $y=x$

graph{(y-(x^3-1)/(x^2-6))(y-x)=0 [-28.86, 28.86, -14.43, 14.45]}

How do you find the asymptotes for f(x) =(x^3-1)/(x^2-6)f(x)=x3−1x2−6f(x) =(x^3-1)/(x^2-6)?