How do you find the slant asymptote of #y=(x^3-4x^2+2x-5)/(x^2+2)#?

1 Answer
Dec 5, 2015

Answer:

#y = x - 4#

Explanation:

First we will expand the expression for #y# by dividing #x^3 - 4x^2 + 2x - 5# through by #x^2 + 2#, using polynomial long division or synthetic division, whichever you prefer. This will yield something that looks like

#y = x - 4 + 3/(x^2+2)#

#y = x - 4# is the equation of the slant asymptote. For large values of #x#, aka when #x# approaches #+- infty#, #3/(x^2+2)# becomes negligible and #y# approaches #x - 4#.

However #y# will never equal #x - 4#. To justify this, we can assume for the sake of argument that for some #x#, #3/(x^2+2)# will equal zero, thus causing #y# to equal #x - 4#:

#0 = 3/(x^2+2)#

If we multiply through by #x^2 + 2# we obtain quite an absurd condition for this to be true, namely that #3 = 0#. Therefore #y# can never exactly equal #x - 4#.

These are the defining characteristics of an asymptote and therefore #x - 4# is this plane curve's slant asymptote.