Find the Domain of y=-(x-12)²+2: Analyzing Input Values

To solve this problem, follow these steps:

• Step 1: Find the roots of the function. Set $y = 0 = -\left(x-12\right)^2 + 2$.

• Step 2: Rearrange and solve for $x$: $\begin{aligned} -\left(x-12\right)^2 + 2 &= 0 \\ \left(x-12\right)^2 &= 2 \end{aligned}$ Solving gives $x - 12 = \pm \sqrt{2}$, resulting in roots $x = 12 + \sqrt{2}$ and $x = 12 - \sqrt{2}$.

• Step 3: Determine the intervals: $\begin{aligned} x < 12 - \sqrt{2} \\ 12 - \sqrt{2} < x < 12 + \sqrt{2} \\ x > 12 + \sqrt{2} \end{aligned}$Step 4: Test each interval to check the sign of $y$: \begin{itemize}

• For $x < 12 - \sqrt{2}$ and $x > 12 + \sqrt{2}$, $(x-12)^2$ becomes larger than 2, so $y= -[(x-12)^2] + 2$ is negative.

• For $12 - \sqrt{2} < x < 12 + \sqrt{2}$, $(x-12)^2$ is less than 2, so $y = -[(x-12)^2] + 2$ is positive.

Thus, the function is negative for $x > 12 + \sqrt{2}$ or $x < 12 - \sqrt{2}$, and positive for $12 - \sqrt{2} < x < 12 + \sqrt{2}$.

Therefore, the positive and negative domains of the function are:

$x > 12+\sqrt{2}$ or $x < 0 : x < 12-\sqrt{2}$

$x > 0 : 12-\sqrt{2} < x < 12+\sqrt{2}$