Let f(x) = x3 - 8x + 13. Then the equation of the tangent line to the graph of f(x) at the point (1,6) is given by y = mx + b, with m = and b =

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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### Question 6

Let \( f(x) = x^3 - 8x + 13 \). Then the equation of the tangent line to the graph of \( f(x) \) at the point \( (1, 6) \) is given by \( y = mx + b \), with

\[ m = \text{[input box]} \]

and

\[ b = \text{[input box]} \]. 

### Explanation

To find the equation of the tangent line, we need to determine the slope \( m \) at the given point by finding \( f'(x) \) and evaluating it at \( x = 1 \). Once we have \( m \), we can use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), to solve for \( b \). The point given is \( (1, 6) \).
Transcribed Image Text:### Question 6 Let \( f(x) = x^3 - 8x + 13 \). Then the equation of the tangent line to the graph of \( f(x) \) at the point \( (1, 6) \) is given by \( y = mx + b \), with \[ m = \text{[input box]} \] and \[ b = \text{[input box]} \]. ### Explanation To find the equation of the tangent line, we need to determine the slope \( m \) at the given point by finding \( f'(x) \) and evaluating it at \( x = 1 \). Once we have \( m \), we can use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), to solve for \( b \). The point given is \( (1, 6) \).
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