Find Intervals of Increase and Decrease for y = -(x-16)²

To solve this problem, we need to determine where the function $y = -(x-16)^2$ is increasing and decreasing.

The function given, $y = -(x-16)^2$, represents a quadratic function with a vertex form of $y = a(x-h)^2 + k$, where $a = -1$, $h = 16$, and $k = 0$. This form shows that the parabola opens downwards because the coefficient $a = -1$ is negative.

The vertex of the parabola, found at $x = 16$, is the point where the function changes its direction. For $x < 16$, since the parabola opens downwards, the function is increasing as it moves toward the vertex. For $x > 16$, the function is decreasing as it moves away from the vertex.

Thus, the intervals are:
- Increasing: $x < 16$
- Decreasing: $x > 16$

The correct solution to the problem, which matches the given answer, is: $\searrow: x > 16 \\ \nearrow: x < 16$.