Find the Domain of y=-(x+10)²+2: Complete Function Analysis

To solve this problem, we start by identifying where the given quadratic function is positive and where it is negative.

• Step 1: Convert the vertex form to solve for $y = 0$. The equation is $-\left(x + 10\right)^2 + 2 = 0$.
• Step 2: Rearrange the equation to find roots:
  $(x+10)^2 = 2$.
  Taking the square root, we have $x + 10 = \pm \sqrt{2}$.
• Step 3: Solve for $x$:
  $x = -10 + \sqrt{2}$ and $x = -10 - \sqrt{2}$.
• Step 4: Analyze the intervals:

The roots divide the number line into intervals. We check these intervals for $y > 0$ and $y < 0$.

• For $y > 0$: The function is a downward-opening parabola, hence positive between its roots: $-10-\sqrt{2} < x < -10+\sqrt{2}$.
• For $y < 0$: The function is negative outside the interval where it is positive, giving $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Therefore, for the positive domain $x > 0$, we have the interval $-10-\sqrt{2} < x < -10+\sqrt{2}$. For the negative domain, it is when $x < 0$ such that $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Thus, the correct solution choice is:

$x > 0 : -10-\sqrt{2} < x < -10+\sqrt{2}$

$x > -10+\sqrt{2}$ or $x < 0 : x < -10-\sqrt{2}$