How do you find the vertical, horizontal and slant asymptotes of: $f (x) = \frac{x^{3}}{x^{2} - 1}$ $f(x)= x^3 / (x^2-1)$ ?

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Answer 1 · 2016-10-21T21:12:19

The vertical asymptotes are $x = 1$ $x=1$ and $x = - 1$ $x=-1$
The slant asymptote is $y = x$ $y=x$

Firstly the domain of $f (x)$ $f(x)$

is $RR-(1,-1)$
since we cannot divide by zero
$x-1!=0$ and $x+1!=0$
so $x!=-1$ and $x!=-1$

So $x=1$ and $x=-1$ are vertical asymptotes

A clant asymptote exists only if the degree of the numerator is greater than that of the numerator

We can do a long division

$f(x)=x^3/(x^2-1)=x+x/(x^2-1)$

So the slant asymptote is $y=x$

How do you find the vertical, horizontal and slant asymptotes of: f(x)= x^3 / (x^2-1)f(x)=x3x2−1f(x)= x^3 / (x^2-1)?