Domain Analysis: Solving (x-6)² + 8 Function Boundaries

To solve this problem, we will explore the behavior of the quadratic function $y = (x-6)^2 + 8$.

The function is in vertex form, $y = (x-h)^2 + k$, where the vertex of the parabola is at $(h, k) = (6, 8)$. The parabola opens upwards because the squared term, $(x-6)^2$, has a positive coefficient (which is 1).

Given this upward-opening parabola, the minimum value of $y$ is $8$, which occurs when $x = 6$. As a result, the quadratic expression $y = (x-6)^2 + 8$ will always yield non-negative values, actually, specifically, it will always yield positive values $y \geq 8$ across its entire domain of real numbers. Therefore, there are no negative values for $y$ in the range of this function, as the minimum bound itself is positive.

Thus, the analysis tells us:

• Negative domain $x < 0$: None, meaning no part of the range $y$ becomes negative.
• Positive domain $x > 0$: All $x$, indicating that $y$ is always positive or zero, overriding negative conditions.

Therefore, the solution for the domains is:

$x < 0$: none

$x > 0$: all $x$