Find the Domain of y=-(x-4)²+1: Positive and Negative Analysis

To solve this problem, we need to find where $y = -\left(x-4\right)^2+1$ is positive or negative.

Step 1: Set $y = 0$ to find the roots.

$-\left(x-4\right)^2+1 = 0$

Solve for $x$:

$-(x-4)^2 = -1$ $(x-4)^2 = 1$

Take the square root of both sides:

$x-4 = \pm 1$

This gives the roots:

$x = 4 \pm 1$

Thus, the roots are $x = 5$ and $x = 3$.

Step 2: Analyze intervals defined by roots.

Intervals are $(-\infty, 3)$, $(3, 5)$, and $(5, \infty)$.

Check sign of $y$ in these intervals:

• For $x < 3$: Choose $x = 0$. Then, $y = -\left(0-4\right)^2+1 = -16 + 1 = -15$. So, $y < 0$.
• For $3 < x < 5$: Choose $x = 4$. Then, $y = -\left(4-4\right)^2+1 = 1$. So, $y > 0$.
• For $x > 5$: Choose $x = 6$. Then, $y = -\left(6-4\right)^2+1 = -4 + 1 = -3$. So, $y < 0$.

The function $y = -\left(x-4\right)^2+1$ is positive for $3 < x < 5$ and negative for $x < 3$ or $x > 5$.

Thus, the positive domain is $3 < x < 5$ and the negative domain is $x > 5$ or $x < 3$.

The correct choices align with the intervals found, which are:

$x > 5$ or $x < 0 : x < 3$

$x > 0 : 3 < x < 5$