To determine which equivalent form of the equation $y = x^2 - 6x - 16$ includes the $x$- and $y$-coordinates of the vertex as constants, let's convert it into vertex form, which is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

Step 1: Start with the original equation

$y = x^2 - 6x - 16$

Step 2: Complete the square

To complete the square for the $x^2 - 6x$ term:

1. Take half the coefficient of $x$ ($-6$) and square it: $\left(\frac{-6}{2}\right)^2 = 9$
2. Add and subtract $9$ to the equation: $y = (x^2 - 6x + 9) - 9 - 16$
3. Simplify: $y = (x - 3)^2 - 25$

Step 3: Identify the vertex

The vertex form is: $y = (x - 3)^2 - 25$ Thus, the vertex is $(3, -25)$.

Step 4: Compare to the answer choices

The correct equivalent form of the equation is: $\boxed{\text{A: } y = (x - 3)^2 - 25}$

Let me know if you want further clarifications or details! Here are some related questions for practice:

1. How do you find the vertex of a parabola in standard form?
2. What does completing the square mean geometrically?
3. Can all quadratic equations be converted into vertex form?
4. How do you determine the axis of symmetry of a parabola?
5. What is the significance of the coefficients in the vertex form equation?

Tip: Completing the square is useful not only for rewriting equations but also for solving quadratic equations and analyzing their properties!