Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.1 - Rates of Change and Tangents to Curves - Exercises 2.1 - Page 46: 7


a. $4$ b. $ y=4x-9$

Work Step by Step

Given $ y=x^2-5, P(2,-1)$ a. Slope=$\frac{(2+h)^2-2^2}{h}=\frac{4h+h^2}{h}=4+h=4$ (as h approach zero) b. tangent line at P: $ y+1=4(x-2)$, which gives $ y=4x-9$
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.