Find the Domain of y=-(x-14)²+8: Analyzing Quadratic Boundaries

To find the positive and negative domains of the function $y = -\left(x-14\right)^2 + 8$, we'll start by identifying the roots of the quadratic equation.

Step 1: Find the roots of the equation:
To find when the function is zero, set $y = 0$:
$-\left(x-14\right)^2 + 8 = 0$.

Step 2: Solve for $x$:
Rearrange the equation:
$-\left(x-14\right)^2 = -8$
$(x-14)^2 = 8$.

Take the square root on both sides:
$x-14 = \pm\sqrt{8}$.
This simplifies to $x - 14 = \pm 2\sqrt{2}$.

Add 14 to both sides to solve for $x$:
$x = 14 \pm 2\sqrt{2}$.
So, the roots are $x = 14 + 2\sqrt{2}$ and $x = 14 - 2\sqrt{2}$.

Step 3: Analyze intervals between roots and outside:
The roots divide the $x$-axis into three intervals: $x < 14 - 2\sqrt{2}$, $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, and $x > 14 + 2\sqrt{2}$.

- For $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, $y > 0$ because points between roots are above the $x$-axis.
- For $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$, $y < 0$ because points outside of roots are below the $x$-axis.

Conclusion:
The positive domain, where $y > 0$, is $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$.
The negative domain, where $y < 0$, is $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$.

Therefore, the solution is:
Positive domain: $x > 0 : 14-2\sqrt{2} < x < 14+2\sqrt{2}$.
Negative domain: $x > 14+2\sqrt{2}$ or $x < 0 : x < 14-2\sqrt{2}$.