Observ your function f(x)=-x/((x-1)(x^2-4))
Asymptotes ( if they exist ) appear on a graph when lim_(x to n)f(x)=oo, n in RR, or if lim_(x to oo)f(x)= ax+b, (a,b) in RR^2
Here, x->1, x->2, x->-2 and x -> oo are good candidates.
( by product, we can see that (x-1)(x^2-4)->0 for x->1, x->2, etc. )
So :
lim_(x to 1)f(x)=lim_(x to 1)-x/((x-1)(x²-4))
Let X=x-1 <=> x=X+1, with X to 0 we can apply basic rules that we know on limits.
Now we have lim_(X to 0)-(X+1)/(X((X+1)²-4)
=lim_(X to 0)-(X+1)/(X(X²+2X-3)
=lim_(X to 0)-(X+1)/(X^3+2X^2-3X
Because we have the divison of polynomials, we only look at the mononial of smallest degree , here, 1 and 3X.
So : lim_(x to 1)f(x)=lim_(X to 0)f(X)=lim_(X to 0)1/(3X)=+-oo
Now let do the exact same thing for x to 2 and x to -2
lim_(x to 2)f(x)=lim_(x to 2)-x/((x-1)(x^2-4))
Let X=x-2
=lim_(X to 0)-(X+2)/((X+1)((X+2)²-4)
=lim_(X to 0)-(X+2)/((X+1)(X²+4X))
=lim_(X to 0)-2/(4X)
= +-oo
I will not detail for x-> -2, we find lim_(X->0)1/(6X)=+-oo
Now let see lim_(x to oo)f(x)
=lim_(x to oo)-x/((x-1)(x²-4))=lim_(x to oo)-x/(x^3-x^2-4x+4)
Now x-> oo, so we only look at the monomials of highest degree.
=lim_(x to oo) -x/(x^3)=lim_(x to oo)-1/x^2=0^-
So they are no asymptotes in oo.
We can easily see it on a graph :
graph{y=-x/((x-1)(x^2-4)) [-4,4,-8,8]}
\0/ Here's our answer !