How to find the asymptotes of #f(x) = (x^2+x-2)/( x^3-3x^2+2x)#?

1 Answer
Dec 22, 2016

Answer:

#f(x)# has vertical asymptotes #x=0# and #x=2#, a horizontal asymptote #y=0# and a hole at #(1, -3)#

Explanation:

Start by factoring both numerator and denominator and cancelling common factors:

#f(x) = (x^2+x-2)/(x^3-3x^2+2x) = ((x+2)color(red)(cancel(color(black)((x-1)))))/(x(x-2)color(red)(cancel(color(black)((x-1))))) = ((x+2))/(x(x-2))#

with exclusion #x != 1#

Notice that when #x=1#, both the numerator and denominator are zero in the given rational function. So #f(1)# is undefined, but the simplified expression is well defined when #x=1#:

#(color(blue)(1)+2)/(color(blue)(1)(color(blue)(1)-2)) = 3/(-1) = -3#

So #f(x)# has a hole (removable singularity) at #(1, -3)#

The remaining values of #x# at which the denominator is zero are #x=0# and #x=2#. In both of these cases the numerator is non-zero. So there are vertical asymptotes at each of these values of #x#.

Finally, since the degree of the numerator is greater than the denominator, #f(x) -> 0# as #x->+-oo#. That is: #f(x)# has horizontal asymptote #y=0#.

It has no slant (oblique) asymptotes. Such slant asymptotes can only occur if the degree of the numerator is #1# greater than the denominator.

Here's a graph of #f(x)# with the vertical asymptotes...

graph{(y-(x+2)/(x(x-2)))(0.9999x-2)(0.9999x+0.0001y) = 0 [-8.71, 11.29, -5.76, 4.24]}