How do you graph #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<=3#.

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

x-intercepts ( y = 0 ) : #+-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#.