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

Dec 5, 2015

$y = x - 4$

#### Explanation:

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

$y = x - 4 + \frac{3}{{x}^{2} + 2}$

$y = x - 4$ is the equation of the slant asymptote. For large values of $x$, aka when $x$ approaches $\pm \infty$, $\frac{3}{{x}^{2} + 2}$ 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 $x$, $\frac{3}{{x}^{2} + 2}$ will equal zero, thus causing $y$ to equal $x - 4$:

$0 = \frac{3}{{x}^{2} + 2}$

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.