The given equation is:

$y = -(x - 6)^2 - 1$

This is a quadratic equation representing a parabola. Here's a breakdown of its key features:

1. Vertex Form: The equation is already in the vertex form of a parabola, which is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

2. Vertex: In this equation, $h = 6$ and $k = -1$. Therefore, the vertex of the parabola is at $(6, -1)$.

3. Direction: The coefficient of the squared term is negative ($a = -1$), so the parabola opens downward.

4. Axis of Symmetry: The axis of symmetry is a vertical line passing through the vertex, which is $x = 6$.

5. Y-intercept: To find the y-intercept, set $x = 0$ and solve for $y$: $y = -(0 - 6)^2 - 1 = -(36) - 1 = -37$ So, the y-intercept is $(0, -37)$.

Would you like more details on any of these points, or do you have any questions?

Here are five related questions to explore further:

1. What is the range of the function $y = -(x - 6)^2 - 1$?
2. How do you find the x-intercepts of this parabola?
3. What is the domain of any quadratic function?
4. How would the graph change if the equation were $y = (x - 6)^2 - 1$ instead?
5. Can you derive the standard form of a quadratic equation from the vertex form?

Tip: Remember, the vertex form of a parabola is handy for quickly identifying the vertex and the direction in which the parabola opens.