Find the Domain of y=-(x-1/3)²+4: Complete Function Analysis

To find the positive and negative domains of the function $y = -\left(x - \frac{1}{3}\right)^2 + 4$, we start by determining where the function crosses the x-axis. This happens where $y = 0$.

Set $y = 0$ to get:

$-\left(x - \frac{1}{3}\right)^2 + 4 = 0$

Solving for $x$:

$\left( x - \frac{1}{3} \right)^2 = 4$ $x - \frac{1}{3} = \pm 2$ $x = \frac{1}{3} \pm 2$

This gives:

Positive root (for $+2$):

$x = \frac{1}{3} + 2 = \frac{7}{3}$

Negative root (for $-2$):

$x = \frac{1}{3} - 2 = -\frac{5}{3}$

The x-intercepts are $x = \frac{7}{3}$ and $x = -\frac{5}{3}$.

Since the quadratic opens downward (as $a = -1$), the graph is above the x-axis between these roots and below outside this interval.

Therefore, the function is positive for:

$x \in \left(-\frac{5}{3}, \frac{7}{3}\right)$

And negative for:

$x > \frac{7}{3}$ or $x < -\frac{5}{3}$

Thus, the positive domain is:

$x > 0 : -\frac{5}{3} < x < \frac{7}{3}$

And the negative domain is:

$x > \frac{7}{3}$ or $x < 0 : x < -\frac{5}{3}$

The correct choice is: