Calculate Area Under y=-(x-3)²: Negative Quadratic Function

To determine the positive area of the function $y = -(x-3)^2$, we start by examining the properties of the given function:

• The function $y = -(x-3)^2$ is a quadratic in the form $y = -a(x-p)^2$ where $a = 1$ and $p = 3$.
• This represents a downward opening parabola with a vertex at $(3, 0)$.

The function is defined as the negative of a square, so $y$ will always be less than or equal to zero. In equation form, $-(x-3)^2 \leq 0$. We solve for the conditions under which this function could be positive:

• For the expression $-(x-3)^2$ to be greater than zero, $-(x-3)^2 > 0$, is impossible because the squared term $(x-3)^2$ is always non-negative and its negative is thus always non-positive.
• At the vertex $x = 3$, $-(x-3)^2 = 0$, the function is exactly zero.

This analysis reveals that the function does not achieve positive values in any part of its domain.

Therefore, there is no positive area for the function $y = -(x-3)^2$.