Find an equation of the tangent line to the curve at the given point. y = In(x² - 4x + 1), (4,0) y =

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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Q5. Please find the equation of the line 

**Instruction:**

Find an equation of the tangent line to the curve at the given point.

**Equation:**

\[ y = \ln(x^2 - 4x + 1) \]

**Given Point:** 

\[ (4, 0) \]

**Solution Box:**

\[ y = \] 

In this task, we are asked to find the equation of the tangent line to the curve defined by \( y = \ln(x^2 - 4x + 1) \) specifically at the point (4, 0). The equation of the tangent line will be in the form of \( y = mx + b \), where \( m \) is the slope of the tangent line at the given point and \( b \) is the y-intercept.
Transcribed Image Text:**Instruction:** Find an equation of the tangent line to the curve at the given point. **Equation:** \[ y = \ln(x^2 - 4x + 1) \] **Given Point:** \[ (4, 0) \] **Solution Box:** \[ y = \] In this task, we are asked to find the equation of the tangent line to the curve defined by \( y = \ln(x^2 - 4x + 1) \) specifically at the point (4, 0). The equation of the tangent line will be in the form of \( y = mx + b \), where \( m \) is the slope of the tangent line at the given point and \( b \) is the y-intercept.
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