y² - 4x - 2y = 7
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Converting and Graphing a Quadratic Equation**
Given the quadratic equation:
\[ y^2 - 4x - 2y = 7 \]
**Objective:**
- Convert the equation to its standard form.
- Graph the equation.
**Step-by-Step Solution:**
1. **Rearrange the equation:** Group the terms involving \(y\) together.
\[ y^2 - 2y - 4x = 7 \]
2. **Complete the square for the \(y\) terms:**
To complete the square:
- Focus on \(y^2 - 2y\).
- Take half of the coefficient of \(y\) (which is \(-2\)), square it, and add & subtract it inside the equation.
\[ y^2 - 2y + 1 - 1 - 4x = 7 \]
\[ (y-1)^2 - 1 - 4x = 7 \]
3. **Further rearrange:**
\[ (y-1)^2 - 4x - 1 = 7 \]
\[ (y-1)^2 - 4x = 8 \]
\[ (y-1)^2 = 4x + 8 \]
To write it in the standard form of a parabola:
\[ (y-1)^2 = 4(x + 2) \]
**Interpreting the Standard Form:**
- The equation \((y - k)^2 = 4p(x - h)\) represents a parabola that opens horizontally.
- Here, \( (h, k) = (-2, 1) \).
- \( 4p = 4 \) implies \( p = 1 \).
So, the parabola opens to the right with the vertex at \((-2, 1)\).
**Graphing the Equation:**
1. **Vertex:** Locate the vertex at \((-2, 1)\).
2. **Direction:** Since \( p > 0 \), the graph opens to the right.
3. **Plot Points:** Plot additional points by substituting values of \( x \) or \( y \).
By plotting the vertex and considering the nature and direction of the parabola, you can sketch the graph accordingly.
**Diagram:**
(Include a graph with the parabola opening rightwards, vertex at \((-2, 1)\),](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8654127-239b-48d1-a891-9934e15b427b%2Fafa13bd1-462b-4898-bec0-ff5320fa1c72%2Fpbfx2mf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Converting and Graphing a Quadratic Equation**
Given the quadratic equation:
\[ y^2 - 4x - 2y = 7 \]
**Objective:**
- Convert the equation to its standard form.
- Graph the equation.
**Step-by-Step Solution:**
1. **Rearrange the equation:** Group the terms involving \(y\) together.
\[ y^2 - 2y - 4x = 7 \]
2. **Complete the square for the \(y\) terms:**
To complete the square:
- Focus on \(y^2 - 2y\).
- Take half of the coefficient of \(y\) (which is \(-2\)), square it, and add & subtract it inside the equation.
\[ y^2 - 2y + 1 - 1 - 4x = 7 \]
\[ (y-1)^2 - 1 - 4x = 7 \]
3. **Further rearrange:**
\[ (y-1)^2 - 4x - 1 = 7 \]
\[ (y-1)^2 - 4x = 8 \]
\[ (y-1)^2 = 4x + 8 \]
To write it in the standard form of a parabola:
\[ (y-1)^2 = 4(x + 2) \]
**Interpreting the Standard Form:**
- The equation \((y - k)^2 = 4p(x - h)\) represents a parabola that opens horizontally.
- Here, \( (h, k) = (-2, 1) \).
- \( 4p = 4 \) implies \( p = 1 \).
So, the parabola opens to the right with the vertex at \((-2, 1)\).
**Graphing the Equation:**
1. **Vertex:** Locate the vertex at \((-2, 1)\).
2. **Direction:** Since \( p > 0 \), the graph opens to the right.
3. **Plot Points:** Plot additional points by substituting values of \( x \) or \( y \).
By plotting the vertex and considering the nature and direction of the parabola, you can sketch the graph accordingly.
**Diagram:**
(Include a graph with the parabola opening rightwards, vertex at \((-2, 1)\),
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