How do you sketch the general shape of #f(x)=-x^4+3x^2-2x-4# using end behavior?

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Answer 1 · 2016-12-05T10:28:48

y-intercept ( x = 0 ) : -4. Cuts x-axis at x = -1 and again between x = -2

and x = -1. As #x to +-oo, y to - oo#. See illustrative graph, for shape.

#y = -x^4(1-3/x^2+2/x^3+4/x^4) to -oo, as x to +-oo#.

Sign changing of a functions y and y'', in an interval, reveals the

presence of zero(s) of the function, within the interval.

#y''=-12x^2+6 =9, to x = +-1/sqrt2 and y'''# is not 0.

So, the points of inflexion ( tangent crossing curve ) are at

#x = +-1/sqrt2#,.

graph{y+x^4-3x^2+2x+4=0 [-20.18, 20.18, -10.1, 10.07]}

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