Find the Domain: y=-(x+10)² - 4 Quadratic Function Analysis

To solve the problem, we first analyze the quadratic function $y = -\left(x + 10\right)^2 - 4$.

Step 1: Identify the vertex.

The function is in vertex form $y = a(x - h)^2 + k$. Here, $a = -1$, $h = -10$, and $k = -4$. Therefore, the vertex is $(-10, -4)$.

Step 2: Determine the direction of the parabola.

Since $a = -1$, the parabola opens downwards. This means the function can only take on either negative values or zero as it cannot have a maximum (i.e., no positive y-values).

Step 3: Analyze the domain of positivity and negativity.

Because the parabola opens downwards and its vertex is the highest point at $(-10, -4)$, all y-values are negative.

Step 4: Determine intersections with the x-axis.

To check for intersections with the x-axis where y = 0, solve: $-\left(x + 10\right)^2 - 4 = 0$.

Rearranging gives $-\left(x + 10\right)^2 = 4$,

which implies $(x + 10)^2 = -4$. Since this yields an imaginary number when solving, the graph does not intersect the x-axis; thus, it is never zero.

Conclusion:

Since the function is negative for all x-values, the positive domain is effectively non-existent.

Checking the choices provided, plug in our understanding:

• For $x < 0$, the positive domain is none, as the function doesn't achieve positive values.
• For $x > 0$, the negative domain is all $x$, as determined.

Thus, the correct answer is:

$x < 0 :$ none
$x > 0 :$ all $x$