Answer
.
Work Step by Step
Let us consider the following functions:
$\begin{align}
& {{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}} \\
& {{f}_{2}}\left( x \right)={{x}^{3}} \\
& {{f}_{3}}\left( x \right)={{x}^{3}}+6{{x}^{2}} \\
& {{f}_{4}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x \\
\end{align}$
${{f}_{5}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x+8$
The graph of the function ${{f}_{2}}\left( x \right)={{x}^{3}}$ has the same shape as the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$, but is shifted 2 units toward the left.
The graph of the function ${{f}_{3}}\left( x \right)={{x}^{3}}+6{{x}^{2}}$ is closer to the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$.
The graph of the function ${{f}_{4}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x $ is approaching closer to the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$ in comparison to the graphs of the functions ${{f}_{2}}\left( x \right)={{x}^{3}}$ and ${{f}_{3}}\left( x \right)={{x}^{3}}+6{{x}^{2}}$.
And the graph of the function ${{f}_{5}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x+8$ coincides with the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$ . That is, the graphs of the two functions are the same.
So, the graphs of the functions ${{f}_{2}}\left( x \right)={{x}^{3}},{{f}_{3}}\left( x \right)={{x}^{3}}+6{{x}^{2}},{{f}_{4}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x $ are approaching to the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$, while the graph of a function ${{f}_{5}}\left( x \right)={{x}^{3}}+6{{x}^{2}}+12x+8$ is the same as the graph of the function ${{f}_{1}}\left( x \right)={{\left( x+2 \right)}^{3}}$.
