The given equation of the hyperbola is:

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

Step 1: Recognizing the General Form of a Hyperbola

This equation is in the standard form for a hyperbola centered at $(h, k)$:

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

For the given equation, comparing it to the general form, we can identify the following values:

• The center is $(h, k) = (0, -2)$.
• $a^2 = 9$, so $a = 3$.
• $b^2 = 16$, so $b = 4$.

Step 2: Finding the Center

• Center: $(0, -2)$

Step 3: Finding the Vertices

The vertices are located at $(h \pm a, k)$ for a horizontal hyperbola (since the $x^2$-term is positive):

• Vertices: $(0 + 3, -2) = (3, -2)$ and $(0 - 3, -2) = (-3, -2)$

Step 4: Finding the Foci

The foci are located at $(h \pm c, k)$, where $c$ is given by $c^2 = a^2 + b^2$: $c^2 = 9 + 16 = 25 \quad \Rightarrow \quad c = 5$

• Foci: $(0 + 5, -2) = (5, -2)$ and $(0 - 5, -2) = (-5, -2)$

Step 5: Equations of Asymptotes

The asymptotes for a horizontal hyperbola have the equations: $y - k = \pm \frac{b}{a}(x - h)$ Substitute the known values of $h = 0$, $k = -2$, $a = 3$, and $b = 4$: $y + 2 = \pm \frac{4}{3}x$ Thus, the equations of the asymptotes are: $y + 2 = \frac{4}{3}x \quad \text{and} \quad y + 2 = -\frac{4}{3}x$ or equivalently: $y = \frac{4}{3}x - 2 \quad \text{and} \quad y = -\frac{4}{3}x - 2$

Summary:

• Center: $(0, -2)$
• Vertices: $(3, -2)$ and $(-3, -2)$
• Foci: $(5, -2)$ and $(-5, -2)$
• Equations of Asymptotes: $y = \frac{4}{3}x - 2$ and $y = -\frac{4}{3}x - 2$

If you'd like more details on how to graph or any other aspect of hyperbolas, feel free to ask!

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Follow-up Questions:

1. How can you graph a hyperbola using the center, vertices, and asymptotes?
2. How do you derive the equation of the asymptotes for a hyperbola?
3. What happens to the shape of the hyperbola if $a$ and $b$ values are swapped?
4. Can you explain the role of the foci in the geometric properties of the hyperbola?
5. How do you convert the equation of a hyperbola into its general form?

Tip: When graphing a hyperbola, start by plotting the center, vertices, and drawing the asymptotes, which guide the shape of the curve.