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

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Answer 1 · 2016-11-23T16:26:03

graph{y+x^3+4x^2-3=0 [-20, 20, -10, 10]}

Sum of the coefficients in #f(-x)# is #1-4+3=0#.

So, x+1 is a factor.

The other factors are -(x-a)(x-b), where a and b are

#(-3+-sqrt 21)/2#. And so,

#f(x)=-(x+1)(x-a)(x-b)# =0,

for #x = -1, -3-7913 and 0.7913#, nearly.

As #x to oo, y to -oo and# as #x to -oo, y to oo#

#f'=-3x^2-8x=0#, when x = 0 and #-8/3#. These give local (f'' < 0 )

maximum f = 3 and local ( f'' > 0 ) minimum # f = -6.48#, nearly.

The inserted graph gives the shape with all these features