Find Domain of y=-(x-12)²-4: Negative Quadratic Function Analysis

The given quadratic function is $y = -\left(x-12\right)^2 - 4$. This function is in vertex form $y = a(x-h)^2 + k$, with $a = -1$, $h = 12$, and $k = -4$. Because $a < 0$, the parabola opens downwards.

To find when $y \geq 0$ (positive domain) and $y \leq 0$ (negative domain), we start by identifying where the function is zero, the x-intercepts. Set $y = 0$:

$-\left(x-12\right)^2 - 4 = 0$

Solving for $x$, isolate the squared term:

$-\left(x-12\right)^2 = 4$

$(x-12)^2 = -4$

No real roots exist because $(x-12)^2$ cannot equal a negative number. Thus, the parabola does not intersect the x-axis, meaning it is entirely below it.

Therefore, the function is negative for all $x$. There are no positive values for $y$.

The positive domain $x > 0$ has no points since the graph is always negative; the negative domain is the entire set of real numbers.

Thus, the correct positive and negative domains are:

$x < 0 :$ none

$x > 0 :$ all $x$