Find the Domain of y=(x-1)²+5: Analyzing Positive and Negative Inputs

To solve this problem, we need to analyze the function $y = (x-1)^2 + 5$, which is a quadratic in vertex form.

Step 1: Identify the Vertex and Orientation
The function is given as $y = (x-1)^2 + 5$, which is in the form $y = a(x-h)^2 + k$. Here, $h = 1$ and $k = 5$, meaning the vertex of the parabola is at $(1, 5)$. Because $a = 1$ (which is positive), the parabola opens upwards.

Step 2: Determine the Minimum Value of $y$
Since the parabola opens upwards, the minimum value of $y$ occurs at the vertex. At the vertex $(1, 5)$, the value of $y$ is 5.

Step 3: Analyze Positive and Negative Values of $y$
The minimum value of $y$ is 5, which indicates that $y$ is always greater than zero. Thus, for all real values of $x$, $y$ remains positive.

Conclusion:
Since the function $y = (x-1)^2 + 5$ has no negative values and is always positive:

$x < 0 :$ none

$x > 0 :$ all $x$

Therefore, the positive and negative domains of the function are:

$x < 0 :$ none

$x > 0 :$ all $x$