.  Given the equation of the parabola (y-1)^2  = 32(x+6). What is the value of the y-coordinate of the vertex?

The given equation of the parabola is \((y - 1)^2 = 32(x + 6)\).

In this equation, the standard form of a horizontally oriented parabola is:

\[
(y - k)^2 = 4p(x - h)
\]

Here, \((h, k)\) represents the vertex of the parabola.

To find the y-coordinate of the vertex, we look at the term \((y - 1)\). Comparing this with the standard form, we can see that \(k = 1\).

Thus, the y-coordinate of the vertex is:

\[
\boxed{1}
\]

Would you like further details or have any other questions? Here are some related questions:

1. What is the x-coordinate of the vertex?
2. How do you find the directrix of this parabola?
3. What is the equation of the axis of symmetry?
4. How do you calculate the distance between the vertex and focus?
5. How does the orientation of the parabola affect its graph?

**Tip**: For horizontally oriented parabolas, the y-coordinate of the vertex remains constant along the axis of symmetry.

Discover how to find the y-coordinate of the vertex of the parabola (y-1)^2 = 32(x+6) with a step-by-step explanation. This problem involves understanding parabolas and their vertices, specifically using the standard form of a horizontally oriented parabola. Suitable for high school students in Grades 9-12.

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