The numerator is
#2x^3+3x^2+x=x(2x^2+3x+1)#
#=x(2x+1)(x+1)#
The denominator is
#6x^2+x-1=(3x-1)(2x+1)#
The function is
#f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#
#=(xcancel(2x+1)(x+1))/((3x-1)cancel(2x+1))#
#=(x(x+1))/(3x-1)#
As the denominator must de #!=0#, therefore
#3x-1!=0#, #=>#, #x!=1/3#
The vertical asymptote is #x=1/3#
There is no horizontal asymptote, as
#lim_(x->oo)f(x)=lim_(x->oo)(x^2)/(3x)=lim_(x->oo)x/3=+oo#
#lim_(x->-oo)f(x)=lim_(x->-oo)(x^2)/(3x)=lim_(x->-oo)x/3=-oo#
As the degree of the numerator is #># the degree of the denominator, there is a slant asymptote.
Perform a long division
#color(white)(aaaa)##x^2+x##color(white)(aaaa)##|##3x-1#
#color(white)(aaaa)##x^2-1/3x##color(white)(aaaa)##|##1/3x+4/9#
#color(white)(aaaaa)##0+4/3x#
#color(white)(aaaaaaa)##+4/3x+0#
#color(white)(aaaaaaa)##+4/3x-4/9#
#color(white)(aaaaaaaaa)##+0+4/9#
Therefore,
#f(x)=(x^2+x)/(3x-1)=(1/3x+4/9)+(4/9)/(3x-1)#
#lim_(x->+oo)(f(x)-(1/3x+4/9))=lim_(x->+oo)(4/9)/(3x-1)=0^+#
#lim_(x->-oo)(f(x)-(1/3x+4/9))=lim_(x->-oo)(4/9)/(3x-1)=0^-#
The slant asymptote is
#y=1/3x+4/9#
graph{(y-(x(x+1))/(3x-1))(y-1/3x-4/9)=0 [-4.382, 4.386, -2.19, 2.193]}
