1. The problem is to understand the inequality $y > -(x-2)^2 + 3$ and describe the region it represents. 2. This inequality involves a quadratic function in vertex form: $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex. 3. Here, $a = -1$, $h = 2$, and $k = 3$, so the parabola opens downward with vertex at $(2,3)$. 4. The inequality $y > -(x-2)^2 + 3$ means we are looking for all points $(x,y)$ where $y$ is greater than the parabola's value. 5. Graphically, this represents the region above the parabola $y = -(x-2)^2 + 3$. 6. The parabola's vertex is the maximum point since $a < 0$. 7. To find intercepts: - $y$-intercept: set $x=0$, then $y = -(0-2)^2 + 3 = -4 + 3 = -1$. - $x$-intercepts: set $y=0$, solve $0 = -(x-2)^2 + 3$ which gives $(x-2)^2 = 3$, so $x = 2 \pm \sqrt{3}$. 8. The solution set is all points above this parabola, not including the parabola itself since the inequality is strict ($>$). Final answer: The inequality $y > -(x-2)^2 + 3$ describes the region above the parabola with vertex at $(2,3)$ opening downward.