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

1 Answer
Dec 5, 2015

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.