Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - F.2 Assess Your Understanding - Page 16: 13

Answer

$(-2,2)$ is not on the graph. $(0,2)$ and $(\sqrt2,\sqrt2)$ are on the graph

Work Step by Step

The points that are on the graph, must satisfy the equation. Therefore we have to see for all three of the points if they satisfy the given equation: $x^2+y^2=4$ For the first point: Left side: $x^2+y^2=0^2+2^2=4$ Right side: $4$ $4=4$ The point is on the graph. For the second point: Left side: $x^2+y^2=(-2)^2+2^2=8$ Right side: $4$ $8\ne4$ The point is not on the graph. For the third point: Left side: $x^2+y^2=\sqrt2^2+\sqrt22^2=4$ Right side: $4$ $4=4$ The point is on the graph.
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