Answer
$y\geq-\frac{1}{2}(x+3)^2+2$
Work Step by Step
The vertex and the two points of the $x$ intercepts are clearly visible from the graph. The vertex and one of the points of the $x$ intercepts can be used to find an equation of the parabola.
Vertex: $(-3,2)$
$x$-intercepts: $(-5,0),(-1,0)$
Hence \begin{equation}
\begin{aligned}
f(x) & =a(x-h)^2+k \\
& =a(x+3)^2+2.
\end{aligned}
\end{equation} We determine $a$: \begin{equation}
\begin{aligned}
f(-5) & =a(-5+3)^2+2 \\
0 & =4 a+2 \\
-2 & =4 a \\
-\frac{1}{2} & =a.
\end{aligned}
\end{equation} The equation of the parabola is. \begin{equation}
f(x)=-\frac{1}{2}(x+3)^2+2.
\end{equation} The graph of the quadratic function is sketched as a solid curve because there is a "equal to" part.
Now we have to find the inequality sign. We take a test point, for example $(0,0)$. The solution is the region which contains this point. We have:
$$\begin{aligned}
y&=0\\
-\frac{1}{2}(x+3)^2+2&=-\frac{1}{2}(0+3)^2+2=-\frac{5}{2}\end{aligned}
$$ As $0\geq -\frac{5}{2}$, we choose the inequality sign, as "greater or equal to" ($\geq$).
The inequality is: $$y\geq -\frac{1}{2}(x+3)^2+2.$$
