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 tangent line to the curve \( y = 5 - 2x - 3x^2 \) at the point \( (1, 0) \).
**Solution:**
1. **Find the derivative of the function:**
The given function is \( y = 5 - 2x - 3x^2 \).
To find the derivative, use the power rule:
\[
\frac{dy}{dx} = -2 - 6x
\]
2. **Evaluate the derivative at the given point:**
The point of tangency is \( (1, 0) \).
Substitute \( x = 1 \) into the derivative:
\[
\frac{dy}{dx} \bigg|_{x=1} = -2 - 6(1) = -8
\]
Thus, the slope (\(m\)) of the tangent line at \( (1, 0) \) is \(-8\).
3. **Use the point-slope form of a line:**
The point-slope form is \( y - y_1 = m(x - x_1) \).
Substitute \( m = -8 \) and the point \( (1, 0) \):
\[
y - 0 = -8(x - 1)
\]
Simplify the equation:
\[
y = -8x + 8
\]
**Conclusion:**
The equation of the tangent line to the curve at the specified point is \( y = -8x + 8 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F021aca7b-3c6d-4d3c-bfc7-736350e47cc8%2F4f05b805-e00c-40cd-a5a2-a3d87d3eea58%2Fbgfdyg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation of the tangent line to the curve \( y = 5 - 2x - 3x^2 \) at the point \( (1, 0) \).
**Solution:**
1. **Find the derivative of the function:**
The given function is \( y = 5 - 2x - 3x^2 \).
To find the derivative, use the power rule:
\[
\frac{dy}{dx} = -2 - 6x
\]
2. **Evaluate the derivative at the given point:**
The point of tangency is \( (1, 0) \).
Substitute \( x = 1 \) into the derivative:
\[
\frac{dy}{dx} \bigg|_{x=1} = -2 - 6(1) = -8
\]
Thus, the slope (\(m\)) of the tangent line at \( (1, 0) \) is \(-8\).
3. **Use the point-slope form of a line:**
The point-slope form is \( y - y_1 = m(x - x_1) \).
Substitute \( m = -8 \) and the point \( (1, 0) \):
\[
y - 0 = -8(x - 1)
\]
Simplify the equation:
\[
y = -8x + 8
\]
**Conclusion:**
The equation of the tangent line to the curve at the specified point is \( y = -8x + 8 \).
Expert Solution
Step 1
We will find the equation of the tangent line to the curve at the given point as following.
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