Find the Domain of (x-2 1/9)² + 5/6: Positive and Negative Regions

To solve this problem, we'll follow these steps:

• Step 1: Recognize that the given quadratic function is in vertex form $y = (x - h)^2 + k$, where $h = 2\frac{1}{9}$ and $k = \frac{5}{6}$.
• Step 2: Identify that the squared term $(x - 2\frac{1}{9})^2$ is always non-negative for any real $x$.
• Step 3: Note that the smallest value that the squared term can obtain is 0, which happens when $x = 2\frac{1}{9}$. Therefore, the smallest value of the whole function $y$ is $\frac{5}{6}$, which is positive. Thus, $y$ is never negative.
• Step 4: Conclude by identifying the domains: $y < 0$ has no solutions and $y > 0$ for all $x$.

After considering the nature of the quadratic function:

Since $y$ cannot be negative, the negative domain is none, which means there are no values where $y < 0$.

On the other hand, for all $x$, $y$ is positive because the minimum value $y$ can take is the constant term $\frac{5}{6}$, which is positive.

Thus, the solution is:

$x < 0 :$ none

$x > 0 :$ for all $x$