Answer
The zeros are $x=0$ with multiplicity 1 and $x=-2$ with multiplicity 2. At $x=0$, the graph will cross the $x\text{-axis}$ and at $x=-2$, the graph will touch the $x\text{-axis}$ and turn around.
Work Step by Step
For zeros, let $f\left( x \right)=0$.
That is,
$\begin{align}
& {{x}^{3}}+4{{x}^{2}}+4x=0 \\
& x\left( {{x}^{2}}+4x+4 \right)=0 \\
& x{{\left( x+2 \right)}^{2}}=0.
\end{align}$
Solving further we get,
$x=0,\ -2$
For multiplicity,
$x=0$, with a multiplicity 1, because of $x$.
$x=-2$, with a multiplicity 2, because of ${{\left( x+2 \right)}^{2}}$.
The multiplicity of $x=0$ is odd; the graph will cross the $x\text{-axis}$.
The multiplicity of $x=-2$ is even; the graph will touch the $x\text{-axis}$ and turn around.
