Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 5

Answer

$y=-8x+12$

Work Step by Step

$y=4x-3x^2, (2,-4)$ $y'=4-6x$ Plug in x-value. $y'=4-6(2)$ $y'=-8$ This is the slope of the tangent line to the curve at the given point. The question asks for an equation of the tangent line. We can use the point slope form. $y-y_{1}=m(x-x_{1})$ $y+4=-8(x-2)$ We can then rewrite this in standard form. $y=-8x+12$
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