Answer
Please see the graph.
Work Step by Step
$y > (x+3)^2 +2$
$y= (x+3)^2 +2$
$y=(x+3)(x+3) +2$
$y=x*x+x*3+x*3+3*3+2$
$y=x^2+6x+9+2$
$y=x^2+6x+11$
$a=1$, $b=6$, $c=11$
$x=-b/2a$
$x=-6/2*1$
$x=-6/2$
$x=-3$
The axis of symmetry (and the vertex) of the graph is at $x=-3$. We pick the point $(-3,5)$ to see what side of the line to shade.
$y > (x+3)^2 +2$
$5 > (-3+3)^2 +2$
$5 > 0^2+2$
$5 > 0+2$
$5 >2$ (true, so we shade the side of the line with this point)