Analyzing the Quadratic Function y = -x² - 4x - 4: Finding Positive and Negative Domains

To find the positive and negative domains of the quadratic function $y = -x^2 - 4x - 4$, we begin by solving for its roots.

Step 1: Calculate the discriminant.
The function is in standard form: $y = ax^2 + bx + c$ with $a = -1$, $b = -4$, $c = -4$.
The discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(-1)(-4) = 16 - 16 = 0$.

Step 2: Find the roots using the quadratic formula.
The quadratic formula is $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.
Since $\Delta = 0$, we have a double root at $x = \frac{-(-4)}{2(-1)} = -2$. Thus, the root is $x = -2$.

Step 3: Determine the sign of $y$ without further roots.
For a quadratic $ax^2 + bx + c$ with $a < 0$, the parabola opens downward. Thus, it will only be positive between the roots if distinct or negative if the root is unique, which, in this case, is at $x = -2$.
$y < 0$ for $x < -2$ and $x > -2$. Since the vertex $(x = -2)$ coincides with the root, this implies $y = 0$ only at $x = -2$.

Step 4: Determine positive and negative domains.
Since the parabola does not exist for positive or zero intervals beyond the vertex and the only root, we conclude:
- For $x < 0$ (all valid $x$ where function changes), $y < 0$ except at $x = -2$.
- For $x > 0$, the function does not cross the x-axis, and $y$ remains negative.

Thus, the final answer is as follows:
$x < 0 : x\ne-2$ and for $x > 0 :$ none.