Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 52

Answer

$y = 2x - 2 + 3 = 2x + 1$

Work Step by Step

first of all lets find the derivative of y: $y = 2x^{3}-x^{2} + 2$ $y' = (2x^{3}-x^{2} + 2)'= (2x^{3})'-(x^{2})' + (2)'=6x^{2}-2x$ So the slope of the tangent line at $(1, 3)$ is (x = 1, and calculate the derivative) $6\times1^{2}-2\times1=6-4=2$ We use the point-slope form to write an equation of the tangent line at $(1,3)$ $y - 3 = 2 (x-1)$ $y = 2x - 2 + 3 = 2x + 1$
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.