Answer
a. $0$
b. $ y=0$
Work Step by Step
Given $ y=x^3-3x^2+4, P(2,0)$
a. slope=$\frac{(2+h)^3-3(2+h)^2-2^3+3\times2^2}{h}=\frac{12h+6h^2+h^3-12h-3h^2}{h}=3h+h^2=0$ (let $h$ be zero)
b. tangent line: $ y=0$
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 47: 14
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