The curve below has a horizontal tangent line at the point (4, – 1) and at one other point. Find the coordinates of the second point where the curve has a horizontal tangent line. 4x? – 32x + 16y² + 96y = – 144 x-coordinate of second point = y-coordinate of second point =
The curve below has a horizontal tangent line at the point (4, – 1) and at one other point. Find the coordinates of the second point where the curve has a horizontal tangent line. 4x? – 32x + 16y² + 96y = – 144 x-coordinate of second point = y-coordinate of second point =
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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![**Finding Points Where the Curve has a Horizontal Tangent Line**
The curve described by the equation \(4x^2 - 32x + 16y^2 + 96y = -144\) has a horizontal tangent line at the point \((4, -1)\) and one other point. To determine the coordinates of the second point where the curve has a horizontal tangent line, follow the steps outlined below.
**Given Equation:**
\[4x^2 - 32x + 16y^2 + 96y = -144\]
**Solution Steps:**
1. **Identify the First Point:**
- There is a point given \((4,-1)\) where the curve has a horizontal tangent line.
2. **Determine the Necessary Calculations:**
- To find horizontal tangent lines on the curve, you need to find points where \(\frac{dy}{dx} = 0\).
3. **Simplify the Given Equation:**
- Rearrange and simplify the given equation if necessary to isolate the terms involving \(x\) and \(y\).
**Graphical Representation:**
- There are no graphs or diagrams provided with the original text, so a graphical interpretation or plotting might help further visually identify where the horizontal tangent lines occur.
**Finding the x and y Coordinates of the Second Point:**
- Use implicit differentiation to find \(\frac{dy}{dx}\) and solve for the conditions where it equals zero.
- Substitute back into the original equation to find the coordinates of the second point.
**Answers Input Fields:**
- **x-coordinate of the second point =** [Text Box]
- **y-coordinate of the second point =** [Text Box]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a9572e1-cd0e-48a9-8ed8-4d34eb3e96b9%2F3a8c7109-93ac-41e0-8822-9be0565e45bd%2Fvg1k0xa_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding Points Where the Curve has a Horizontal Tangent Line**
The curve described by the equation \(4x^2 - 32x + 16y^2 + 96y = -144\) has a horizontal tangent line at the point \((4, -1)\) and one other point. To determine the coordinates of the second point where the curve has a horizontal tangent line, follow the steps outlined below.
**Given Equation:**
\[4x^2 - 32x + 16y^2 + 96y = -144\]
**Solution Steps:**
1. **Identify the First Point:**
- There is a point given \((4,-1)\) where the curve has a horizontal tangent line.
2. **Determine the Necessary Calculations:**
- To find horizontal tangent lines on the curve, you need to find points where \(\frac{dy}{dx} = 0\).
3. **Simplify the Given Equation:**
- Rearrange and simplify the given equation if necessary to isolate the terms involving \(x\) and \(y\).
**Graphical Representation:**
- There are no graphs or diagrams provided with the original text, so a graphical interpretation or plotting might help further visually identify where the horizontal tangent lines occur.
**Finding the x and y Coordinates of the Second Point:**
- Use implicit differentiation to find \(\frac{dy}{dx}\) and solve for the conditions where it equals zero.
- Substitute back into the original equation to find the coordinates of the second point.
**Answers Input Fields:**
- **x-coordinate of the second point =** [Text Box]
- **y-coordinate of the second point =** [Text Box]
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