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The given equation for the hyperbola is:

\[
\frac{(y+2)^2}{16} - \frac{(x-2)^2}{9} = 1
\]

### Standard Form:
This is the standard form of a hyperbola, where the y-term comes first, indicating it is a vertical hyperbola. The general form of a vertical hyperbola is:

\[
\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1
\]

Comparing the equation with the standard form, we can identify the following:

- \( h = 2 \)
- \( k = -2 \)
- \( a^2 = 16 \) → \( a = 4 \)
- \( b^2 = 9 \) → \( b = 3 \)

### Asymptotes:
For a vertical hyperbola, the equations of the asymptotes are given by:

\[
y - k = \pm \frac{a}{b} (x - h)
\]

Substituting the values of \( h \), \( k \), \( a \), and \( b \):

\[
y + 2 = \pm \frac{4}{3}(x - 2)
\]

So, the equations of the asymptotes are:

\[
y + 2 = \frac{4}{3}(x - 2) \quad \text{and} \quad y + 2 = -\frac{4}{3}(x - 2)
\]

These are the asymptotes for the given hyperbola.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you determine if a hyperbola is vertical or horizontal?
2. What are the foci of this hyperbola, and how are they calculated?
3. How would the graph change if the \(x\)-term came first in the equation?
4. What are the vertices of this hyperbola, and how are they found?
5. How do asymptotes help in sketching the graph of a hyperbola?

**Tip:** Always look at the sign of the terms in the equation to quickly determine the orientation of a hyperbola!

Find the asymptote for the hyperbola: \frac{(y+2)^2}{16} - \frac{(x-2)^2}{9} = 1

Learn how to find the asymptotes for the hyperbola \frac{(y+2)^2}{16} - \frac{(x-2)^2}{9} = 1 using the standard hyperbola equation. This problem covers key concepts related to conic sections and asymptote formulas. Suitable for students in Grades 10-12.

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