Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 181: 62

Answer

The other point of intersection is $(\frac{3}{2}, \frac{5}{4})$ We can see the sketch below.
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Work Step by Step

$y = x^2-1$ $\frac{dy}{dx} = 2x$ The slope of the graph at $x = -1$ is: $~~\frac{dy}{dx} =2(-1) = -2$ The slope of the normal line is $~~-(\frac{1}{-2}) = \frac{1}{2}$ We can find the equation of the normal line: $(y-0) = \frac{1}{2}[x-(-1)]$ $y = \frac{1}{2}x+\frac{1}{2}$ We can find the points of intersection: $x^2-1 = \frac{1}{2}x+\frac{1}{2}$ $2x^2-2 = x+1$ $2x^2-x-3 = 0$ $(x+1)(2x-3) = 0$ $x = -1, \frac{3}{2}$ When $x = \frac{3}{2},~~$ then $~~y = (\frac{3}{2})^2-1 = \frac{5}{4}$ The other point of intersection is $(\frac{3}{2}, \frac{5}{4})$ We can see the sketch below.
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