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

The given equation is $y - 4 = -(x - 4)^2$.

First, identify the vertex: the equation is in vertex form $y = -(x - 4)^2+4$ with vertex at $(4, 4)$. The parabola opens downwards because the coefficient of $(x-4)^2$ is negative.

Next, find the x-intercepts by setting $y = 0$:

$0 = -(x - 4)^2 + 4$

$(x - 4)^2 = 4$

Taking the square root of both sides gives $x - 4 = \pm 2$.

So, the solutions are $x = 6$ and $x = 2$.

The parabola is negative between these x-intercepts. Since it opens downwards, the function is negative outside the interval $[2, 6]$, i.e., for $x < 2$ or $x > 6$.

Thus, the negative area of the parabola is for $x < 2$ or $x > 6$, matching choice 2.