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

1 Answer
Feb 1, 2017

See graph and explanation.

Explanation:

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

$y = f \left(x\right) = - {x}^{4} + 3 \le 3$.

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

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

y-intercept ( x = 0 ) : 3

$y ' = - 4 {x}^{3} = 0 , a t x = \left(0 , 3\right)$

At this point, $y ' ' = y ' ' ' = 0 \mathmr{and} y ' ' ' ' = - 24 \ne = 0$.

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

As $x \to \pm \infty , y \to - \infty$.