How do you graph $f (x) = - x^{4} + 3$ $f(x)=-x^4+3$ using zeros and end behavior?

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Answer 1 · 2017-02-01T12:10:59

See graph and explanation.

graph{y+x^4-3=0 [-10, 10, -5, 5]}

$y = f (x) = - x^{4} + 3 \leq 3$ $y = f(x)= -x^4+3<=3$ .

f(-x)=f(x). So, the graph is symmetrical about y-axis.

x-intercepts ( y = 0 ) : $\pm \sqrt{\sqrt{3}} = \pm 1.3161$ $+-sqrtsqrt3=+-1.3161$ , nearly.

y-intercept ( x = 0 ) : 3

$y'=-4x^3=0, at x = (0, 3)$

At this point, $y''=y'''=0 and y''''=-24 ne =0$ .

So, (0, 3) is a POI (point of inflexion ).

As $x to +-oo, y to -oo$ .

How do you graph f(x)=-x^4+3f(x)=−x4+3f(x)=-x^4+3 using zeros and end behavior?