Answer
See image:
Work Step by Step
Completing the square,
$\begin{aligned}f(x)&=x^{2}+4x+2\\&=\left(x^{2}+4x+4\right)+2-4\\&=(x+2)^{2}-2\end{aligned}$
Let $f_{1}(x)=x^{2}$.
Then, $f(x)=f_{1}(x+2)-2$.
Starting with the graph of $f_{1}(x)=x^{2}$ (black-dashed),
shift it left 2 units to obtain the graph of $f_{1}(x+2)$ (green dashed),
and then lower it 2 units down to obtain $f_{1}(x+2)-2=f(x)$ (red).
