Find the Domain: (x-9⅓)² - 4 Function Analysis

To find the positive and negative domains of the function $y = \left(x - 9\frac{1}{3}\right)^2 - 4$, we will solve for when $y = 0$ and analyze the intervals:

First, set the function equal to zero:

• $\left(x - 9\frac{1}{3}\right)^2 - 4 = 0$
• $\left(x - 9\frac{1}{3}\right)^2 = 4$
• Take the square root of both sides: $x - 9\frac{1}{3} = \pm 2$

Calculate the roots:

• For $x - 9\frac{1}{3} = 2$, $x = 11\frac{1}{3}$
• For $x - 9\frac{1}{3} = -2$, $x = 7\frac{1}{3}$

The roots are $x = 7\frac{1}{3}$ and $x = 11\frac{1}{3}$. These are the points where the function crosses the x-axis.

To determine the positive and negative domains, consider the following intervals:

• When $x < 7\frac{1}{3}$, the expression $(x - 9\frac{1}{3})^2$ is greater than 4, making $y > 0$, so the function is positive.
• Between $x = 7\frac{1}{3}$ and $x = 11\frac{1}{3}$, the expression is less than 4, making $y < 0$, so the function is negative.
• When $x > 11\frac{1}{3}$, the expression goes back to being greater than 4, making $y > 0$, so the function is positive.

Therefore, the domains are:

• Positive: $x > 11\frac{1}{3}$ or $x < 7\frac{1}{3}$
• Negative: $7\frac{1}{3} < x < 11\frac{1}{3}$

The correct choice based on this analysis is:

$x < 0 : 7\frac{1}{3} < x < 11\frac{1}{3}$

$x > 11\frac{1}{3}$ or $x > 0 : x < 7\frac{1}{3}$