Domain Analysis: Solving (x-6)² - 3 Function Boundaries

To find the positive and negative domains of the function $y = (x-6)^2 - 3$, follow these steps:

• Step 1: Set $y = 0$ to find the roots of the equation. This gives us $(x-6)^2 - 3 = 0$.
• Step 2: Add 3 to both sides to simplify, resulting in $(x-6)^2 = 3$.
• Step 3: Take the square root of both sides. We have two solutions: $x - 6 = \sqrt{3}$ and $x - 6 = -\sqrt{3}$.
• Step 4: Solve for $x$ from each equation:
• For $x - 6 = \sqrt{3}$, $x = 6 + \sqrt{3}$.
• For $x - 6 = -\sqrt{3}$, $x = 6 - \sqrt{3}$.
• Step 5: Determine the intervals:
• Interval 1: $x < 6 - \sqrt{3}$, insert a test point to determine sign.
• Interval 2: $6 - \sqrt{3} < x < 6 + \sqrt{3}$, insert a test point to determine sign.
• Interval 3: $x > 6 + \sqrt{3}$, insert a test point to determine sign.
• Step 6: Based on the intervals:
• For $x < 6 - \sqrt{3}$ and $x > 6 + \sqrt{3}$, $y > 0$.
• For $6 - \sqrt{3} < x < 6 + \sqrt{3}$, $y < 0$.

The positive domains are: $x < 6 - \sqrt{3}$ or $x > 6 + \sqrt{3}$.

The negative domain is: $6 - \sqrt{3} < x < 6 + \sqrt{3}$.

The correct answer to the problem is:

$x > 6+\sqrt{3}$ or $x < 0 : x < 6-\sqrt{3}$

$x < 0 : 6-\sqrt{3} < x < 6+\sqrt{3}$