Hyperbola identification and graphing worksheet with solutions.

Problem Analysis:

The task involves identifying the vertices, foci, and asymptotes of hyperbolas given their equations, and then sketching their graphs. Let's solve each part step by step.

---

Problem 1:

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

#### Step 1: Identify the type of hyperbola
The equation is in the standard form:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\]
This represents a horizontal transverse axis hyperbola.

#### Step 2: Extract values of \(a\) and \(b\)
From the equation:
\[
a^2 = 9 \implies a = 3
\]
\[
b^2 = 25 \implies b = 5
\]

#### Step 3: Find the vertices
For a horizontal transverse axis hyperbola, the vertices are located at \((\pm a, 0)\):
\[
\text{Vertices: } (3, 0) \text{ and } (-3, 0)
\]

#### Step 4: Find the foci
The distance from the center to each focus is given by \(c\), where:
\[
c = \sqrt{a^2 + b^2}
\]
Substitute \(a^2 = 9\) and \(b^2 = 25\):
\[
c = \sqrt{9 + 25} = \sqrt{34}
\]
The foci are located at \((\pm c, 0)\):
\[
\text{Foci: } (\sqrt{34}, 0) \text{ and } (-\sqrt{34}, 0)
\]
Approximating \(\sqrt{34} \approx 5.83\):
\[
\text{Foci: } (\approx 5.8, 0) \text{ and } (\approx -5.8, 0)
\]

#### Step 5: Find the asymptotes
The equations of the asymptotes for a horizontal transverse axis hyperbola are:
\[
y = \pm \frac{b}{a} x
\]
Substitute \(a = 3\) and \(b = 5\):
\[
y = \pm \frac{5}{3} x
\]

#### Step 6: Sketch the graph
- Center: \((0, 0)\)
- Vertices: \((3, 0)\) and \((-3, 0)\)
- Foci: \((\approx 5.8, 0)\) and \((\approx -5.8, 0)\)
- Asymptotes: \(y = \frac{5}{3} x\) and \(y = -\frac{5}{3} x\)

Sketch the hyperbola with these features.

---

Problem 2:

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

#### Step 1: Identify the type of hyperbola
The equation is in the standard form:
\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\]
This represents a vertical transverse axis hyperbola with center \((h, k) = (3, -4)\).

#### Step 2: Extract values of \(a\) and \(b\)
From the equation:
\[
a^2 = 1 \implies a = 1
\]
\[
b^2 = 1 \implies b = 1
\]

#### Step 3: Find the vertices
For a vertical transverse axis hyperbola, the vertices are located at \((h, k \pm a)\):
\[
\text{Vertices: } (3, -4 + 1) \text{ and } (3, -4 - 1)
\]
\[
\text{Vertices: } (3, -3) \text{ and } (3, -5)
\]

#### Step 4: Find the foci
The distance from the center to each focus is given by \(c\), where:
\[
c = \sqrt{a^2 + b^2}
\]
Substitute \(a^2 = 1\) and \(b^2 = 1\):
\[
c = \sqrt{1 + 1} = \sqrt{2}
\]
The foci are located at \((h, k \pm c)\):
\[
\text{Foci: } (3, -4 + \sqrt{2}) \text{ and } (3, -4 - \sqrt{2})
\]
Approximating \(\sqrt{2} \approx 1.41\):
\[
\text{Foci: } (3, \approx -2.6) \text{ and } (3, \approx -5.4)
\]

#### Step 5: Find the asymptotes
The equations of the asymptotes for a vertical transverse axis hyperbola are:
\[
y - k = \pm \frac{a}{b} (x - h)
\]
Substitute \(a = 1\), \(b = 1\), \(h = 3\), and \(k = -4\):
\[
y + 4 = \pm \frac{1}{1} (x - 3)
\]
\[
y + 4 = \pm (x - 3)
\]
\[
y = x - 7 \quad \text{and} \quad y = -x - 1
\]

#### Step 6: Sketch the graph
- Center: \((3, -4)\)
- Vertices: \((3, -3)\) and \((3, -5)\)
- Foci: \((3, \approx -2.6)\) and \((3, \approx -5.4)\)
- Asymptotes: \(y = x - 7\) and \(y = -x - 1\)

Sketch the hyperbola with these features.

---

Final Answers:

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

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

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