# How do you graph f(x)=x^4-4 using zeros and end behavior?

Oct 30, 2016

Find the zeros, end behavior and y intercept as described below.

#### Explanation:

Graph $f \left(x\right) = {x}^{4} - 4$ using zeros and end behavior.

To find the zeros, factor the polynomial.

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

Factor again.

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

Setting each factor equal to zero and solving gives:

$x = - \sqrt{2}$, $x = \sqrt{2}$ and $x = \pm \sqrt{2} i$

The only real zeros are $\sqrt{2}$ and $- \sqrt{2}$. Each has a multiplicity of $\textcolor{red}{1}$ because the exponent on each factor is $\textcolor{red}{1}$. An odd multiplicity means the graph crosses (or cuts through) the $x$ axis at the zeros/x-intercepts. The $x$ intercepts are $\left(- \sqrt{2} , 0\right)$ and $\left(\sqrt{2} , 0\right)$ which are approximately $\left(\pm 1.414 , 0\right)$.

To find the end behavior, examine the degree and leading coefficient of the original polynomial.

$f \left(x\right) = \textcolor{b l u e}{1} {x}^{\textcolor{v i o \le t}{4}} - 4$

The degree is $\textcolor{v i o \le t}{4}$ and the leading coefficient is $\textcolor{b l u e}{1}$.

An even degree with a positive leading coefficient indicates that as$x \rightarrow \infty$ and $x \rightarrow - \infty$, $f \left(x\right) \rightarrow \infty$. In other words, the "ends" of the graph both point "up".

It is also helpful to find the $y$ intercept. Setting $x = 0$ gives
$y = {0}^{4} - 4 = - 4$. The $y$ intercept is $\left(0 , - 4\right)$

The graph is shown below.