1. The problem is to find the range of the function $g(x) = -2(x+3)^2 + 8$. 2. This is a quadratic function in vertex form $g(x) = a(x-h)^2 + k$, where the vertex is at $(h, k)$. 3. Here, $a = -2$, $h = -3$, and $k = 8$. Since $a < 0$, the parabola opens downward, meaning the vertex is a maximum point. 4. The maximum value of $g(x)$ is the $y$-coordinate of the vertex, which is $8$. 5. The term $(x+3)^2$ is always non-negative, so the smallest value it can take is $0$. 6. As $(x+3)^2$ increases, $-2(x+3)^2$ becomes more negative, decreasing $g(x)$. 7. Therefore, the range of $g(x)$ is all values less than or equal to $8$. 8. In interval notation, the range is $$(-\infty, 8].$$