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

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