Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 5

Answer

The equation of the tangent line $l$ at point $A(2,-4)$ is $$(l):y=-8x+12$$

Work Step by Step

$$y=f(x)=4x-3x^2$$ Given point $A(2,-4)$ According to definition, the slope of the tangent line $l$ at point $A(2, -4)$ is $$m_l=\lim\limits_{x\to2}\frac{f(x)-f(2)}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{(4x-3x^2)-(-4)}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{-3x^2+4x+4}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{-3(x-2)(x+2/3)}{x-2}$$$$m_l=\lim\limits_{x\to2}[-3(x+\frac{2}{3})]$$$$m_l=-3\times(2+\frac{2}{3})=-8$$ Therefore, the equation of the tangent line $l$ would have the following form: $$(l): y=-8x+b$$ Since $l$ passes through point $A(2,-4)$, we have $$-8\times2+b=-4$$$$-16+b=-4$$$$b=12$$ In conclusion, the equation of the tangent line $l$ at point $A(2,-4)$ is $$(l):y=-8x+12$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.