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

To solve where the function $y = (x-8)^2 - 1$ is increasing, let's examine its form:

• The function is in vertex form: $y = (x-8)^2 - 1$ where the vertex is located at the point $(8, -1)$.
• The coefficient of $(x-8)^2$ is positive ($a = 1$), indicating the parabola opens upwards.
• For an upwards-opening parabola, the function will be decreasing to the left of the vertex and increasing to the right.
• Thus, the function is increasing for values of $x$ that are greater than the x-coordinate of the vertex.

Therefore, the interval where the function is increasing is $(x > 8)$.

Thus, the correct answer is choice 4: $x > 8$.