# How do you find the end behavior of y= -1/(x^3+2)?

May 26, 2015

A pragmatic answer is to try substituting large positive and negative values for $x$ and see what happens.

For example, if we substitute $x = - 100$ then

$y = - \frac{1}{{\left(- 100\right)}^{3} + 2} = - \frac{1}{- 1000000 + 2} = - \frac{1}{-} 999998 = \frac{1}{999998}$

... a small positive number.

If we substitute $x = 100$ then

$y = - \left(\frac{1}{{100}^{3} + 2}\right) = - \frac{1}{1000000 + 2} = - \frac{1}{1000002}$

... a small negative number.

We can reasonably deduce that $y \to 0 +$ as $x \to - \infty$
and $y \to 0 -$ as $x \to \infty$.

Notice how the ${x}^{3}$ term becomes dominant for large positive or negative values of $x$ and the constant term becomes irrelevant.

If you have a polynomial factor you can usually ignore the lower order terms when evaluating end behaviour. The exceptions are when you have multiple terms that cause the high order terms to cancel out, leaving the lower order ones to dominate.