Find Decreasing Intervals: Analyzing y = -(x-12)² - 4 in Vertex Form

The function $y = -(x-12)^2 - 4$ is given in vertex form $y = a(x-h)^2 + k$ where $a = -1$, $h = 12$, and $k = -4$. This tells us the vertex of the parabola is at $(12, -4)$. Since $a = -1$ is negative, the parabola opens downward.

In such a parabola, the function is increasing to the left of the vertex and decreasing to the right. The axis of symmetry is $x = 12$. To the left of $x = 12$, the function increases, and to the right of $x = 12$, the function decreases.

Therefore, the function is decreasing when $x > 12$.

Thus, the interval where the function $y = -(x-12)^2 - 4$ is decreasing is for $x > 12$.

The correct answer to this problem is: $x > 12$.