Graph the hyperbola by writing the equation in standard form. Label the center, vertices and sketch the asymptotes. x2 – 4y2 12x 16y + 16 = 0 %3D -
Graph the hyperbola by writing the equation in standard form. Label the center, vertices and sketch the asymptotes. x2 – 4y2 12x 16y + 16 = 0 %3D -
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Graphing Hyperbolas in Standard Form
### Problem Statement
Graph the hyperbola by writing the equation in standard form. Label the center, vertices, and sketch the asymptotes.
Given equation:
\[ x^2 - 4y^2 - 12x - 16y + 16 = 0 \]
### Instructions:
1. **Rewrite the Equation in Standard Form:**
- Completing the square for \(x\): \(x^2 - 12x\)
\[
x^2 - 12x + 36 \quad (\text{adding} \ 36)
\]
- Completing the square for \(y\): \(-4(y^2 + 4y)\)
\[
-4(y^2 + 4y + 4) = -4(y + 2)^2 \quad (\text{adding} \ 16 \ \text{but since it's multiplied by -4, we actually subtract} 64)
\]
- Rewrite the equation:
\[
(x - 6)^2 - 4(y + 2)^2 = 64 \ (\text{transferring added/subtracted constants appropriately})
\]
- Divide by 64 to get it in standard form:
\[
\frac{(x - 6)^2}{64} - \frac{(y + 2)^2}{16} = 1
\]
2. **Identifying Key Components:**
- Center: \((6, -2)\)
- Transverse axis along \(x\):
- Vertices: \((6 + 8, -2) \ \text{and} \ (6 - 8, -2)\) or \((14, -2) \ \text{and} \ (-2, -2)\)
- Asymptotes:
\[
y + 2 = \pm \frac{4}{8}(x - 6) \quad (\text{or more simplified as} \ y + 2 = \pm \frac{1}{2}(x - 6))
\]
### Graph Explanation:
- A grid with axes labeled from -10 to 10 both on x and y axes.
- The hyperbola’s center \((6, -2)\) will be marked.
- Vertices](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c941f29-ddef-4885-9d8b-99e8246804cb%2F34496278-420b-4242-ba1f-07af38480f56%2Fx020bh.jpeg&w=3840&q=75)
Transcribed Image Text:## Graphing Hyperbolas in Standard Form
### Problem Statement
Graph the hyperbola by writing the equation in standard form. Label the center, vertices, and sketch the asymptotes.
Given equation:
\[ x^2 - 4y^2 - 12x - 16y + 16 = 0 \]
### Instructions:
1. **Rewrite the Equation in Standard Form:**
- Completing the square for \(x\): \(x^2 - 12x\)
\[
x^2 - 12x + 36 \quad (\text{adding} \ 36)
\]
- Completing the square for \(y\): \(-4(y^2 + 4y)\)
\[
-4(y^2 + 4y + 4) = -4(y + 2)^2 \quad (\text{adding} \ 16 \ \text{but since it's multiplied by -4, we actually subtract} 64)
\]
- Rewrite the equation:
\[
(x - 6)^2 - 4(y + 2)^2 = 64 \ (\text{transferring added/subtracted constants appropriately})
\]
- Divide by 64 to get it in standard form:
\[
\frac{(x - 6)^2}{64} - \frac{(y + 2)^2}{16} = 1
\]
2. **Identifying Key Components:**
- Center: \((6, -2)\)
- Transverse axis along \(x\):
- Vertices: \((6 + 8, -2) \ \text{and} \ (6 - 8, -2)\) or \((14, -2) \ \text{and} \ (-2, -2)\)
- Asymptotes:
\[
y + 2 = \pm \frac{4}{8}(x - 6) \quad (\text{or more simplified as} \ y + 2 = \pm \frac{1}{2}(x - 6))
\]
### Graph Explanation:
- A grid with axes labeled from -10 to 10 both on x and y axes.
- The hyperbola’s center \((6, -2)\) will be marked.
- Vertices
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