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

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

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Answer 1 · 2016-10-30T02:18:54

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

Explanation:

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

To find the zeros, factor the polynomial.

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

Factor again.

$f (x) = {(x + \sqrt{2})}^{1} {(x - \sqrt{2})}^{1} (x^{2} + 2)$ $f(x)=(x+sqrt2)^color(red)1(x-sqrt2)^color(red)1(x^2+2)$

Setting each factor equal to zero and solving gives:

$x = - \sqrt{2}$ $x=-sqrt2$ , $x = \sqrt{2}$ $x=sqrt2$ and $x = \pm \sqrt{2} i$ $x=+-sqrt2i$

The only real zeros are $\sqrt{2}$ $sqrt2$ and $- \sqrt{2}$ $-sqrt2$ . Each has a multiplicity of $1$ $color(red)1$ because the exponent on each factor is $1$ $color(red)1$ . An odd multiplicity means the graph crosses (or cuts through) the $x$ $x$ axis at the zeros/x-intercepts. The $x$ $x$ intercepts are $(- \sqrt{2}, 0)$ $(-sqrt2,0)$ and $(\sqrt{2}, 0)$ $(sqrt2, 0)$ which are approximately $(\pm 1.414, 0)$ $(+-1.414,0)$ .

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

$f (x) = 1 x^{4} - 4$ $f(x)=color(blue)1x^color(violet)4-4$

The degree is $4$ $color(violet)4$ and the leading coefficient is $1$ $color(blue)1$ .

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

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

The graph is shown below.

![desmos.com]( useruploads.socratic.org )

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

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