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Start by graphing the boundary using the corresponding function, \(y = -2(x - 4)^2 - 7\). The inequality is strict, so graph the boundary with a dashed curve.



A test point can be used to determine which region to shade.  Here, \((0,0)\) is used.

Left Side	Right Side
\[\begin{array}{r} y \\  0 \end{array}\]	\[\begin{array}{l} -2(x - 4)^2 - 7\\ -2((0) - 4)^2 - 7 \\ -39    \end{array}\]
\(\hspace{25pt}\)LS \(\gt\) RS

The point \((0,0)\) does not satisfy the inequality, so it and all other points on that side of the boundary are not solutions. Shade the region that does not include the test point.

The point \((5, -5)\) lies outside the solution region, so it is not a solution. This can also be checked algebraically.

Left Side	Right Side
\[\begin{array}{r} y \\  -5 \end{array}\]	\[\begin{array}{l} -2(x - 4)^2 - 7\\ -2((5) - 4)^2 - 7 \\ -9    \end{array}\]
\(\hspace{25pt}\)LS \(\gt\) RS

The point \((5,-5)\) does not satisfy the inequality, so it is not a solution.

Start by writing the equation of the boundary. The vertex is at \((-2, 2)\), and \((1, 5)\) is another point on the curve.

\(\begin{align}
 y &= a\left( {x + 2} \right)^2 + 2 \\
 5 &= a\left( {1 + 2} \right)^2 + 2 \\
 3 &= 9a \\
 \frac{1}{3} &= a \\
 \end{align}\)

The equation of the boundary is \(y = \frac{1}{3}(x + 2)^2 + 2\). The curve is solid, so the inequality is not strict. Use a test point from the solution region to determine the direction of the inequality symbol. The point \((1, 0)\) is in the solution region.

Left Side	Right Side
\[\begin{array}{r} y \\  0 \end{array}\]	\[\begin{array}{l} \frac{1}{3}(x + 2)^2 + 2\\ \frac{1}{3}((1) + 2)^2 + 2 \\ 5    \end{array}\]
\(\hspace{25pt}\)LS \(\lt\) RS

An inequality that represents the graph is \(y \le \frac{1}{3}(x + 2)^2 + 2\).