Find the Domain of y=(x-5)²: Analyzing Positive and Negative Inputs

To determine the positive and negative domains of the function $y = (x - 5)^2$, follow these steps:

• Step 1: Recognize the function is in vertex form: $y = (x - h)^2 + k$, where the vertex is $(h, k) = (5, 0)$.
• Step 2: Observe the properties of a square. The expression $(x - 5)^2$ is always greater than or equal to zero because a square of any real number cannot be negative.
• Step 3: Analyze when the function is zero. The output $y = 0$ when $x - 5 = 0$, or $x = 5$.
• Step 4: Consider where $y$ is positive. For all $x$ except $x = 5$, the square $(x-5)^2 > 0$.
• Step 5: Determine where $y$ is negative. Since a squared term is never negative, there are no values of $x$ for which $y < 0$.

Based on these steps, the positive domain captures all $x$ except where $x = 5$, and the negative domain is nonexistent because the square is always non-negative.

Therefore, the solution is:

$x < 0 :$ none
$x > 0 : x\ne5$