Hyperbola equations and their graphical representations with labeled vertices, foci, and asymptotes.

Problem Analysis:

The task involves identifying the vertices, foci, and asymptotes of hyperbolas given their equations, and then sketching their graphs. The equations provided are in standard forms for hyperbolas.

#### Standard Forms of Hyperbolas:
1. Horizontal Transverse Axis:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\]
- Vertices: \((\pm a, 0)\)
- Foci: \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\)
- Asymptotes: \(y = \pm \frac{b}{a}x\)

2. Vertical Transverse Axis:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]
- Vertices: \((0, \pm a)\)
- Foci: \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\)
- Asymptotes: \(y = \pm \frac{a}{b}x\)

---

Solution for Each Problem:

#### Problem 1:
Equation:
\[
\frac{x^2}{9} - \frac{y^2}{25} = 1
\]

1. Identify \(a\) and \(b\):
- From the equation, \(a^2 = 9\) and \(b^2 = 25\).
- Therefore, \(a = 3\) and \(b = 5\).

2. Vertices:
- Since the transverse axis is horizontal, the vertices are at \((\pm a, 0)\).
- Vertices: \((3, 0)\) and \((-3, 0)\).

3. Foci:
- Calculate \(c\):
\[
c = \sqrt{a^2 + b^2} = \sqrt{9 + 25} = \sqrt{34}
\]
- The foci are at \((\pm c, 0)\).
- Foci: \((\sqrt{34}, 0)\) and \((- \sqrt{34}, 0)\).
- Approximate \(\sqrt{34} \approx 5.8\).
- Foci: \((5.8, 0)\) and \((-5.8, 0)\).

4. Asymptotes:
- The asymptotes for a horizontal hyperbola are given by \(y = \pm \frac{b}{a}x\).
- Here, \(\frac{b}{a} = \frac{5}{3}\).
- Asymptotes: \(y = \frac{5}{3}x\) and \(y = -\frac{5}{3}x\).

5. Sketch the Graph:
- Plot the vertices at \((3, 0)\) and \((-3, 0)\).
- Plot the foci at \((5.8, 0)\) and \((-5.8, 0)\).
- Draw the asymptotes \(y = \frac{5}{3}x\) and \(y = -\frac{5}{3}x\).
- Sketch the hyperbola branches approaching the asymptotes.

#### Problem 2:
Equation:
\[
(y + 4)^2 - (x - 3)^2 = 1
\]

1. Identify \(a\) and \(b\):
- Rewrite the equation in standard form:
\[
\frac{(y + 4)^2}{1} - \frac{(x - 3)^2}{1} = 1
\]
- Here, \(a^2 = 1\) and \(b^2 = 1\).
- Therefore, \(a = 1\) and \(b = 1\).

2. Vertices:
- The center of the hyperbola is \((h, k) = (3, -4)\).
- Since the transverse axis is vertical, the vertices are at \((h, k \pm a)\).
- Vertices: \((3, -4 + 1) = (3, -3)\) and \((3, -4 - 1) = (3, -5)\).

3. Foci:
- Calculate \(c\):
\[
c = \sqrt{a^2 + b^2} = \sqrt{1 + 1} = \sqrt{2}
\]
- The foci are at \((h, k \pm c)\).
- Foci: \((3, -4 + \sqrt{2})\) and \((3, -4 - \sqrt{2})\).
- Approximate \(\sqrt{2} \approx 1.4\).
- Foci: \((3, -2.6)\) and \((3, -5.4)\).

4. Asymptotes:
- The asymptotes for a vertical hyperbola are given by \(y - k = \pm \frac{a}{b}(x - h)\).
- Here, \(\frac{a}{b} = 1\).
- Asymptotes: \(y + 4 = \pm 1(x - 3)\).
- Simplify:
\[
y + 4 = x - 3 \quad \text{or} \quad y = x - 7
\]
\[
y + 4 = -x + 3 \quad \text{or} \quad y = -x - 1
\]
- Asymptotes: \(y = x - 7\) and \(y = -x - 1\).

5. Sketch the Graph:
- Plot the center at \((3, -4)\).
- Plot the vertices at \((3, -3)\) and \((3, -5)\).
- Plot the foci at \((3, -2.6)\) and \((3, -5.4)\).
- Draw the asymptotes \(y = x - 7\) and \(y = -x - 1\).
- Sketch the hyperbola branches approaching the asymptotes.

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Final Answers:

1. For \(\frac{x^2}{9} - \frac{y^2}{25} = 1\):
- Vertices: \((3, 0)\), \((-3, 0)\)
- Foci: \((5.8, 0)\), \((-5.8, 0)\)
- Asymptotes: \(y = \frac{5}{3}x\), \(y = -\frac{5}{3}x\)

2. For \((y + 4)^2 - (x - 3)^2 = 1\):
- Vertices: \((3, -3)\), \((3, -5)\)
- Foci: \((3, -2.6)\), \((3, -5.4)\)
- Asymptotes: \(y = x - 7\), \(y = -x - 1\)

\[
\boxed{
\begin{aligned}
1. & \text{ Vertices: } (3, 0), (-3, 0) \\
& \text{ Foci: } (5.8, 0), (-5.8, 0) \\
& \text{ Asymptotes: } y = \frac{5}{3}x, y = -\frac{5}{3}x \\
2. & \text{ Vertices: } (3, -3), (3, -5) \\
& \text{ Foci: } (3, -2.6), (3, -5.4) \\
& \text{ Asymptotes: } y = x - 7, y = -x - 1
\end{aligned}
}
\]