Find the Domain of (x+5)²-6: Analyzing Positive and Negative Regions

To determine where the function $y = (x + 5)^2 - 6$ is positive and negative, we start by solving the equation:

$(x + 5)^2 - 6 = 0$

Adding 6 to both sides gives:

$(x + 5)^2 = 6$

Taking the square root of both sides, we obtain two solutions:

$x + 5 = \sqrt{6}$ or $x + 5 = -\sqrt{6}$

Solving these, we get:

$x = -5 + \sqrt{6}$ and $x = -5 - \sqrt{6}$

These roots divide the number line into three intervals: $(- \infty, -5 - \sqrt{6})$, $(-5 - \sqrt{6}, -5 + \sqrt{6})$, and $(-5 + \sqrt{6}, \infty)$.

Next, we determine the sign of the function in each interval:

• For $x < -5 - \sqrt{6}$, choose $x = -10$:

$(x + 5)^2 - 6 = ((-10) + 5)^2 - 6 = (-5)^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

• For $-5 - \sqrt{6} < x < -5 + \sqrt{6}$, choose $x = -5$:

$(x + 5)^2 - 6 = ((-5) + 5)^2 - 6 = 0^2 - 6 = -6 < 0$. Therefore, the function is negative.

• For $x > -5 + \sqrt{6}$, choose $x = 0$:

$(x + 5)^2 - 6 = (0 + 5)^2 - 6 = 5^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

Thus, the function is positive on the intervals $x < -5 - \sqrt{6}$ and $x > -5 + \sqrt{6}$, and negative on the interval $-5 - \sqrt{6} < x < -5 + \sqrt{6}$.

Therefore, the positive domain is $x > -5+\sqrt{6}$ or $x > 0 : x < -5-\sqrt{6}$, and the negative domain is $x < 0 : -5-\sqrt{6} < x < -5+\sqrt{6}$.