Answer
$y
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: $(0,-12)$
$x$-intercepts: $(-4,0),(3,0)$
\begin{equation}
\begin{aligned}
& f(x)=a x^2+b x+c \\
& f(0)=c=-12
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
f(x)& =a x^2+b x-12 \\
f(-4)& =16 a-4 b-12=0 \\
f(3) & = 9a+3b-12=0
\end{aligned}
\end{equation} Now solve the system of equations for the values of $a$ and $b$. \begin{equation}
\begin{cases}
16 a-4 b & =12 \\
9 a+3 b & =12
\end{cases}
\end{equation} Solve for $b$ in term of $a$.
\begin{equation}
\begin{aligned}
3 b & =12-9 a \\
b & =\frac{12-9 a}{3} \\
& =4-3 a
\end{aligned}
\end{equation} Substitute for $b$ in the first equation and solve. \begin{equation}
\begin{aligned}
16 a-4(4-3 a) & =12 \\
16 a-16+12 a & =12 \\
28 a & =12+16 \\
28 a & =28 \\
a & =1.
\end{aligned}
\end{equation} and \begin{equation}
\begin{aligned}
b & =4-3(1) \\
& =1.
\end{aligned}
\end{equation} The equation of the parabola is.
$$ f(x) = x^2+x-12.$$ The graph of the quadratic function is sketched as a dashed curve because there is no "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:
$$\begin{aligned}
y&=0\\
x^2+x-12&=0^2+0-12=-12\end{aligned}
$$ As $0>-12$, we choose the opposite inequality sign, which is "less than" ($<$).
The inequality is: $$y
