## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 1 - Review - Exercises - Page 70: 18

#### Answer

When $-2\leq x\leq-1, y=-2x-2$ When $-1\lt x\leq1, y=\sqrt{1-x^2}$

#### Work Step by Step

For $-2\leq x\leq-1,$ $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ $y-2=\frac{0-2}{-1-(-2)}(x-(-2))$ $y=-2x-2$ For $-1\lt x\leq1,$ $x^2+y^2=1, y\geq0$ $y=\sqrt{1-x^2}$

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