Domain Analysis: Finding Valid Inputs for (x-8⅙)² - 1

To find the positive and negative domains of $y=\left(x-8\frac{1}{6}\right)^2-1$, we perform the following steps:

• Step 1: Identify when $y = 0$. Set $\left(x-8\frac{1}{6}\right)^2-1 = 0$, solving gives $\left(x-8\frac{1}{6}\right)^2 = 1$.
• Step 2: Solve for $x$ to get $\left(x-8.1667\right)^2 = 1$. The equation gives $x-8.1667 = \pm 1$ leading to two solutions: $x = 9.1667$ and $x = 7.1667$.
• Step 3: Determine the sign of $y$ in intervals $(-\infty, 7.1667)$, $(7.1667, 9.1667)$, and $(9.1667, \infty)$.
• Step 4: Over the interval $(7.1667, 9.1667)$, $y \lt 0$ because the shifted-square is less than 1, making $(x-8.1667)^2-1 \lt 0$.
• Step 5: Outside this interval, specifically $(-\infty, 7.1667)$ and $(9.1667, \infty)$, $y \gt 0$.

The positive domain is $x \in (-\infty, 7\frac{1}{6}) \cup (9\frac{1}{6}, \infty)$ and the negative domain is $x \in (7\frac{1}{6}, 9\frac{1}{6})$.

Therefore, the solution to the problem is:

$x < 0 : x <7\frac{1}{6} < x < 9\frac{1}{6}$

$x > 9\frac{1}{6}$ or $x > 0 : x <7\frac{1}{6}$