What are the asymptotes of $f(x)=-x/((x-1)(x^2+x))$ ?

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Answer 1 · 2016-12-20T10:37:44

There is a hole at $x=0$
The vertical asymptotes are $x=-1$ and $x=1$
No slant asymptote.
The horizontal asymptote is $y=0$

Let's factorise the denominator

$(x-1)(x^2+x)=x(x-1)(x+1)$

Therefore,

$f(x)=cancelx/(cancelx(x+1)(x-1))$

$=1/((x+1)(x-1))$

There is a hole at $x=0$

The domain of $f(x)$ is $D_f(x)=RR-{-1,1}$

As you cannot divide by $0$ , $x!=-1$ and $x!=1$

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

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

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

The horizontal asymptote is $y=0$

graph{1/((x+1)(x-1)) [-14.24, 14.24, -7.11, 7.14]}

What are the asymptotes of f(x)=-x/((x-1)(x^2+x)) f(x)=-x/((x-1)(x^2+x)) ?