Explanation Given information: f ( t ) = 6 t − t 2 at t = 4 Formula used: The slope-intercept form of the equation is y − y 1 = m ( x − x 1 ) The slope m of the tangent line is m = f ' ( x 0 ) where x 0 is the x-coordinate of the given point. Calculation: f ( t ) = 6 t − t 2 (Given) Differentiating with respect to ‘t’, we get f ' ( t ) = 6 ( 1 ) − ( 2 t ) = 6 − 2 t Substituting t = 4 in the above equation, we get f ' ( 4 ) = 6 − 2 ( 4 ) = 6 − 8 ⇒ m = − 2 Substituting m = − 2 in the equation y − y 1 = m ( x − x 1 ) , we get y − y 1 = − 2 ( x − x 1 ) Substituting t = 4 in f

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ISBN: 9781119275565

Author: Hughes-Hallett

Publisher: WILEY

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Chapter 3.1, Problem 60P

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The equation for the tangent line to the graph of f(t)=6t−t2 at t=4.

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Chapter 3.1, Problem 59P

Chapter 3.1, Problem 61P

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The equation for the tangent line to the graph of f ( t ) = 6 t − t 2 at t = 4.

Solution Summary: The author explains how the equation for the tangent line is y=-2t+16.

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The equation for the tangent line to the graph of f(t)=6t−t2 at t=4.

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Chapter 3 Solutions

The equation for the tangent line to the graph of f ( t ) = 6 t − t 2 at t = 4.

Solution Summary: The author explains how the equation for the tangent line is y=-2t+16.

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To find: The equation for the tangent line to the graph of f(t)=6t−t2 at t=4.

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Chapter 3 Solutions

To find:

The equation for the tangent line to the graph of f(t)=6t−t2 at t=4.