Domain Analysis of y=x²-8x+16: Finding Positive and Negative Regions

To determine the positive and negative domains of the function $y = x^2 - 8x + 16$, follow these steps:

• Step 1: Re-express the quadratic in standard form. Notice that $x^2 - 8x + 16$ can be rewritten as $(x-4)^2$ because it is a perfect square trinomial. This simplifies identification of intercepts.
• Step 2: Determine the vertex and the direction of the parabola. Since $(x - 4)^2$ is a perfect square, it has a minimum at $x = 4$ when $y = 0$. Thus, the vertex is (4, 0), and the parabola opens upwards.
• Step 3: Identify intervals on the x-axis where the function is positive or zero. Since it opens upwards, $y = (x-4)^2 \geq 0$ for all $x$ and is zero exactly at $x = 4$.
• Step 4: Recognize that the function can only be non-positive or zero, never negative, given its parabolic nature with an upward opening.

Therefore, the positive domain of the function is $x > 0$ excluding $x = 4$, and there is no negative domain, consistent with the solution: $x > 0 : x \ne 4$; $x < 0 :$ none.

This conclusion leads us to verify with choice options. The correct option matches this solution, indicating the positive interval where the function remains non-negative and is bounded by $x = 4$ as a minimum.