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 the equation of the tangent line to the curve \( y = (5 - 3x)^2 \) at \( x = -5 \).
**Solution:**
To find the equation of the tangent line, we need to determine the slope of the tangent line at the given point and the y-coordinate of that point.
1. **Differentiate the function:**
The function given is \( y = (5 - 3x)^2 \).
Using the chain rule, the derivative is:
\[
y' = 2(5 - 3x)(-3) = -6(5 - 3x)
\]
2. **Find the slope at \( x = -5 \):**
Substitute \( x = -5 \) into the derivative:
\[
y'(-5) = -6(5 - 3(-5)) = -6(5 + 15) = -6 \times 20 = -120
\]
The slope of the tangent line is \(-120\).
3. **Find the y-coordinate of the point on the curve:**
Substitute \( x = -5 \) into the original function to find the y-coordinate:
\[
y = (5 - 3(-5))^2 = (5 + 15)^2 = 20^2 = 400
\]
The point on the curve is \((-5, 400)\).
4. **Write the equation of the tangent line:**
Use the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = -120 \), \( x_1 = -5 \), and \( y_1 = 400 \).
\[
y - 400 = -120(x + 5)
\]
Distribute and simplify:
\[
y = -120x - 600 + 400
\]
\[
y = -120x - 200
\]
Thus, the equation of the tangent line is \( y = -120x - 200 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8256360-52ce-4e6c-9144-e8e1be652f97%2F78c812b7-e137-4350-8ea2-177eea962519%2F1il0gdn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of the tangent line to the curve \( y = (5 - 3x)^2 \) at \( x = -5 \).
**Solution:**
To find the equation of the tangent line, we need to determine the slope of the tangent line at the given point and the y-coordinate of that point.
1. **Differentiate the function:**
The function given is \( y = (5 - 3x)^2 \).
Using the chain rule, the derivative is:
\[
y' = 2(5 - 3x)(-3) = -6(5 - 3x)
\]
2. **Find the slope at \( x = -5 \):**
Substitute \( x = -5 \) into the derivative:
\[
y'(-5) = -6(5 - 3(-5)) = -6(5 + 15) = -6 \times 20 = -120
\]
The slope of the tangent line is \(-120\).
3. **Find the y-coordinate of the point on the curve:**
Substitute \( x = -5 \) into the original function to find the y-coordinate:
\[
y = (5 - 3(-5))^2 = (5 + 15)^2 = 20^2 = 400
\]
The point on the curve is \((-5, 400)\).
4. **Write the equation of the tangent line:**
Use the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = -120 \), \( x_1 = -5 \), and \( y_1 = 400 \).
\[
y - 400 = -120(x + 5)
\]
Distribute and simplify:
\[
y = -120x - 600 + 400
\]
\[
y = -120x - 200
\]
Thus, the equation of the tangent line is \( y = -120x - 200 \).
Expert Solution
Step 1
To find the equation of the tangent line to the below curve at x = -5.
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