Find Decreasing Intervals: Analyzing y = (x-8)² - 1

The function $y = (x-8)^2 - 1$ is a quadratic equation in vertex form, indicating a parabola. A parabola in this form $y = a(x-h)^2 + k$ has a vertex at $(h, k)$. For this function, the vertex is located at $(8, -1)$.

Since the coefficient of $(x-8)^2$ is positive (specifically, $a = 1$) the parabola opens upwards. The axis of symmetry is the vertical line $x = 8$, around which the parabola is symmetric. This line divides the parabola into sections where it is decreasing and increasing.

To the left of this vertex (for $x < 8$), the function is decreasing. To the right of this vertex (for $x > 8$), the function is increasing. This is because, as we move away from the vertex on an upward-opening parabola's left side, the y-values decrease.

In conclusion, the interval over which the function is decreasing is $x < 8$.