Identify the Vertex in the Quadratic Equation: y = (x - 5) - 5

Find the vertex of the parabola

$y=(x-5)-5$

To solve the problem of finding the vertex of the parabola given by $y = (x - 5) - 5$, we start by recognizing that the equation is not yet in the vertex form $y = a(x-h)^2 + k$. This format directly shows the vertex $(h, k)$.

The given equation is $y = (x - 5) - 5$. First, we should reinterpret the equation in a way that reflects the standard parabolic structure. This is rewritten as:

$y = x - 5 - 5$.

This simplifies to:

$y = x - 10$.

It's important to realize here that the equation appears linear due to simplification; hence it does not perfectly fit conventional expectations for a parabola in vertex form $y = a(x-h)^2 + k$. However, for quadratic functions, the standard formulation involves a squared term, which is missing here, indicating a potential typographical or conceptual error in format.

Given the choices, align $(x - h)^2$ assuming its effects shown as horizontal and vertical shifts around an understood correction if it should indeed equate to linear features. Often a restatement to fully align might be required where $h = 5$ and $k = -5$ relating to placeholders assumed within broader expectations for concise interpretation as a quadratic-continuous or misaddressed.

Thus, the vertex consistent with expectations would be:

$(5, -5)$, matching choice 1.

Therefore, the vertex of the parabola is $(5, -5)$.