How do you find the vertical, horizontal or slant asymptotes for #f(x)=(x^3)/((x-1)^2)#?

1 Answer
Jan 3, 2017

Answer:

The vertical asymptote is #x=1#
No horozontal asym`ptote.
The slant asymptote is #y=x+2#

Explanation:

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

As you cannot divide by #0#, #x!=1#

The vertical asymptote is #x=1#

As the degree of the numerator is #># than the degree of the denominator, we have a slant asymptote.

The denominator is

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

We do a long division

#color(white)(aaaa)##x^3##color(white)(aaaaaaaaaaaaaaa)##∣##x^2-2x+1#

#color(white)(aaaa)##x^3-2x^2+x##color(white)(aaaaaa)##∣##x+2#

#color(white)(aaaaa)##0+2x^2-x#

#color(white)(aaaaaaa)##+2x^2-4x+2#

#color(white)(aaaaaaaaa)##+0+3x-2#

So,

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

Therefore,

#lim_(x->-oo)f(x)-(x+2)=lim_(x->-oo)(3x)/x^2=lim_(x->-oo)3/x=0^(-)#

#lim_(x->+oo)f(x)-(x+2)=lim_(x->+oo)(3x)/x^2=lim_(x->+oo)3/x=0^(+)#

So,

The slant asymptote is #y=x+2#

graph{(y-(x^3/(x-1)^2))(y-x-2)(y-100(x-1))=0 [-27.35, 30.37, -11.67, 17.2]}