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!