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...
Related questions
Question
![**Problem Statement:**
Find an equation of the normal line to the parabola \( y = x^2 - 8x + 4 \) that is parallel to the line \( x - 4y = 4 \).
**Solution:**
- **Step 1:** Determine the slope of the given line.
- Rewrite \( x - 4y = 4 \) in slope-intercept form:
\[
4y = x - 4 \quad \Rightarrow \quad y = \frac{1}{4}x - 1
\]
- The slope of the line is \( \frac{1}{4} \).
- **Step 2:** Find the derivative of the parabola to get the slope of the tangent line at any point \( (x, y) \). The derivative of \( y = x^2 - 8x + 4 \) is:
\[
\frac{dy}{dx} = 2x - 8
\]
- **Step 3:** For the normal line to be parallel to \( x - 4y = 4 \), it must have the same slope as \( \frac{1}{4} \). Since the normal line is perpendicular to the tangent line at the point of tangency, the product of their slopes must be \(-1\):
\[
m_{\text{tangent}} \times \frac{1}{4} = -1 \quad \Rightarrow \quad m_{\text{tangent}} = -4
\]
- **Step 4:** Set the slope of the tangent line \( 2x - 8 \) equal to \(-4\):
\[
2x - 8 = -4 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2
\]
- **Step 5:** Substitute \( x = 2 \) back into the equation of the parabola to find the \( y \)-coordinate:
\[
y = (2)^2 - 8(2) + 4 = 4 - 16 + 4 = -8
\]
The point of tangency is \( (2, -8) \).
- **Step 6:** Use the point-slope form to find the equation of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdbb678aa-6d9d-4962-a0fc-b4c9c9ece909%2F4ce08b64-264b-46d5-adb1-81a54f56afef%2Fausjcsn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation of the normal line to the parabola \( y = x^2 - 8x + 4 \) that is parallel to the line \( x - 4y = 4 \).
**Solution:**
- **Step 1:** Determine the slope of the given line.
- Rewrite \( x - 4y = 4 \) in slope-intercept form:
\[
4y = x - 4 \quad \Rightarrow \quad y = \frac{1}{4}x - 1
\]
- The slope of the line is \( \frac{1}{4} \).
- **Step 2:** Find the derivative of the parabola to get the slope of the tangent line at any point \( (x, y) \). The derivative of \( y = x^2 - 8x + 4 \) is:
\[
\frac{dy}{dx} = 2x - 8
\]
- **Step 3:** For the normal line to be parallel to \( x - 4y = 4 \), it must have the same slope as \( \frac{1}{4} \). Since the normal line is perpendicular to the tangent line at the point of tangency, the product of their slopes must be \(-1\):
\[
m_{\text{tangent}} \times \frac{1}{4} = -1 \quad \Rightarrow \quad m_{\text{tangent}} = -4
\]
- **Step 4:** Set the slope of the tangent line \( 2x - 8 \) equal to \(-4\):
\[
2x - 8 = -4 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2
\]
- **Step 5:** Substitute \( x = 2 \) back into the equation of the parabola to find the \( y \)-coordinate:
\[
y = (2)^2 - 8(2) + 4 = 4 - 16 + 4 = -8
\]
The point of tangency is \( (2, -8) \).
- **Step 6:** Use the point-slope form to find the equation of the
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Step 1: Find the equation of normal line
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