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

Nov 23, 2016

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

Sum of the coefficients in $f \left(- x\right)$ is $1 - 4 + 3 = 0$.

So, x+1 is a factor.

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

$\frac{- 3 \pm \sqrt{21}}{2}$. And so,

$f \left(x\right) = - \left(x + 1\right) \left(x - a\right) \left(x - b\right)$ =0,

for $x = - 1 , - 3 - 7913 \mathmr{and} 0.7913$, nearly.

As $x \to \infty , y \to - \infty \mathmr{and}$ as $x \to - \infty , y \to \infty$

$f ' = - 3 {x}^{2} - 8 x = 0$, when x = 0 and $- \frac{8}{3}$. 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