Answer
a. $12$
b. $ y=12x-16$
Work Step by Step
Given $ y=x^3, P(2,8)$
a. slope=$\frac{(2+h)^3-2^3}{h}=\frac{3\times4h+6h^2+h^3}{h}=12+6h+h^2=12$ (let $h$ be zero)
b. tangent line: $ y-8=12(x-2)$, which is $ y=12x-16$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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: 11
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