College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.2 - Graphs of Functions - 2.2 Exercises: 77

Answer

$f(x)=\sqrt{9-x^{2}}, -3\leq x\leq 3$

Work Step by Step

We are given $y=x^{2}+y^{2}=9=3^2$ This describes a circle with radius 3 centered at (0,0). We solve for $y$: $y^{2}=9-x^{2}$ $y=\pm\sqrt{9-x^{2}}$ We want the top half of the circle, so we take the positive: $y=\sqrt{9-x^{2}}$ Thus the function is: $f(x)=\sqrt{9-x^{2}}, -3\leq x\leq 3$
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