Find the Domain of y=-(x+4)²+6: Analyzing Positive and Negative Inputs

The function given is $y = -\left(x+4\right)^2 + 6$. This is in vertex form $y = a(x-h)^2 + k$ with vertex at $(-4, 6)$.

Step 1: To find the x-values for which the function is positive or negative, set $y = 0$:

$-\left(x+4\right)^2 + 6 = 0$

$\left(x+4\right)^2 = 6$

Step 2: Solve for $x$:

Take the square root of both sides:

$x + 4 = \pm \sqrt{6}$ i.e., $x = -4 \pm \sqrt{6}$

Step 3: Find where the function is positive or negative. The parabola opens downward, so the intervals are:

• Negative domain: $x < -4 - \sqrt{6}$ and $x > -4 + \sqrt{6}$; outside this interval.
• Positive domain: $-4 - \sqrt{6} < x < -4 + \sqrt{6}$; within this interval.

Conclusively:

$x > 0 : -4-\sqrt{26} < x < -4+\sqrt{26}$

$x > -4+\sqrt{26}$ or $x < 0 : x < -4-\sqrt{26}$

Therefore, the solution to this problem is as follows:

For $x > 0$: $-4-\sqrt{26} < x < -4+\sqrt{26}$

For $x < 0$: $x < -4-\sqrt{26}$ and $( x > -4 + \sqrt{26})$