Answer
$ y\leq \frac{3}{5} x^2+\frac{12}{5} x-3$
Work Step by Step
The $y$ intercept and the two points of the $x$ intercepts are clearly visible from the graph. These points can be used to find an equation of the parabola.
$y$- intercept: $(0,-3)$
$x$-intercepts: $(-5,0),(1,0)$
\begin{equation}
\begin{aligned}
& f(x)=a x^2+b x+c \\
& f(0)=c=-3
\end{aligned}
\end{equation} We determine $a$ and $b$:
\begin{equation}
\begin{aligned}
f(x)& =a x^2+b x-3 \\
f(-5)& =25 a-5 b-3=0 \\
f(1) &=a+b-3=0.
\end{aligned}
\end{equation} Now solve the system of equations for the values of $a$ and $b$. \begin{equation}
\begin{cases}
25 a-5 b & =3 \\
a+3 & =3
\end{cases}
\end{equation} Solve for $b$ in term of $a$. \begin{equation}
\begin{aligned}
b & =3-a \\
\end{aligned}
\end{equation} Substitute for $b$ in the first equation and solve. \begin{equation}
\begin{aligned}
25a-5(3-a)& =3 \\
25a+5a-15 & =3 \\
30 a & =3+15 \\
30 a & =18\\
a=\frac{18}{30}\\
a=\frac{3}{5}
\end{aligned}
\end{equation} and \begin{equation}
\begin{aligned}
b & =3-\frac{3}{5} \\
& =\frac{3(5)-3}{5} \\
& =\frac{12}{5}
\end{aligned}
\end{equation} The equation of the parabola is. \begin{equation}
f(x)=\frac{3}{5} x^2+\frac{12}{5} x-3.
\end{equation} The graph of the quadratic function is sketched as a solid curve because there is an "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 does not contain this point. We have: $$\begin{aligned}
y&=0\\
\frac{3}{5} x^2+\frac{12}{5} x-3&=\frac{3}{5} (0^2)+\frac{12}{5}(0)-3\\
&=-3\end{aligned}
$$ As $0\geq -3$, we choose the opposite inequality sign "less or equal to" ($\leq$).
The inequality is: $$y\leq \frac{3}{5} x^2+\frac{12}{5} x-3.$$
