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 !
