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

Aug 3, 2017

${\lim}_{x \to + \infty} f \left(x\right) = + \infty$

${\lim}_{x \to - \infty} f \left(x\right) = + \infty$

#### Explanation:

Given:

$f \left(x\right) = - {x}^{2} \left(1 - 2 x\right) \left(x + 2\right)$

In order to determine the end behaviour, we only need to look at the term of highest degree, i.e. the leading term in standard form. Rather than multiply out the whole expression, we can just multiply the terms of highest degree in each factor to find:

$- {x}^{2} \left(- 2 x\right) \left(x\right) = 2 {x}^{4}$

So the term of highest degree in $f \left(x\right)$ is $2 {x}^{4}$, which is of even degree with positive coefficient.

As a result the end behaviour is:

${\lim}_{x \to + \infty} f \left(x\right) = + \infty$

${\lim}_{x \to - \infty} f \left(x\right) = + \infty$

In general, the possible combinations in a polynomial function are:

• Even degree, positive leading coefficient:
$\textcolor{w h i t e}{}$
${\lim}_{x \to \pm \infty} = + \infty$
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• Odd degree, positive leading coefficient:
$\textcolor{w h i t e}{}$
${\lim}_{x \to + \infty} = + \infty \text{ } {\lim}_{x \to - \infty} = - \infty$
$\textcolor{w h i t e}{}$

• Even degree, negative leading coefficient:
$\textcolor{w h i t e}{}$
${\lim}_{x \to \pm \infty} = - \infty$
$\textcolor{w h i t e}{}$

• Odd degree, negative leading coefficient:
$\textcolor{w h i t e}{}$
${\lim}_{x \to + \infty} = - \infty \text{ } {\lim}_{x \to - \infty} = + \infty$