Calculate Area Under y=(x-6)² + 4: Quadratic Function Problem

We begin by analyzing the given function $y = (x-6)^2 + 4$. This is a parabola in vertex form, with the vertex at the point $(6, 4)$. The expression $(x-6)^2$ represents the square of $(x-6)$, and as a square, it can only be zero or positive, thus $(x-6)^2 \geq 0$.

Adding 4, the smallest value that $y$ can take is therefore $4$, which occurs when $(x-6)^2 = 0$. That means at the vertex, the function achieves its minimum value of $y = 4$, which is clearly positive.

This implies the function $y = (x-6)^2 + 4$ is never negative for any real value of $x$. Hence, the area above the x-axis is positive or non-negative for all values of $x$, which means no restrictions on the domain are necessary based on positivity.

Thus, the solution concludes that the function remains non-negative for all $x$, satisfying the condition of positive area.

Therefore, the correct choice is For all X.