Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 21: 54


$y=2+\sqrt{4-x^2}$ With domain: $-2\leq x \leq2$

Work Step by Step

We need to solve for $y$: $x^2+(y-2)^2=4$ $(y-2)^2=4-x^2$ $y-2=\pm\sqrt{4-x^2}$ $y=2\pm\sqrt{4-x^2}$ Since we need the top half of the circle, we use: $y=2+\sqrt{4-x^2}$ With domain: $-2\leq x \leq2$
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