Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 103: 53

Answer

$(0,-2)$ is on the graph of the equation. $(1,-2)$ is not on the graph of the equation. $(2,-2)$ is on the graph of the equation.

Work Step by Step

$x^{2}+xy+y^{2}=4;$ $(0,-2)$ $,$ $(1,-2)$ $,$ $(2,-2)$ Substitute each point into the equation: $(0,-2)$ $x^{2}+xy+y^{2}=4$ $0^{2}+(0)(-2)+(-2)^{2}=4$ $0+0+4=4$ $4=4$ True Substituting the point $(0,-2)$ into the equation makes it true. This point is on the graph of the equation. $(1,-2)$ $x^{2}+xy+y^{2}=4$ $1^{2}+(1)(-2)+(-2)^{2}=4$ $1-2+4=4$ $3\ne4$ False Substituting the point $(1,-2)$ into the equation makes it false. This point is not on the graph of the equation. $(2,-2)$ $x^{2}+xy+y^{2}=4$ $2^{2}+(2)(-2)+(-2)^{2}=4$ $4-4+4=4$ $4=4$ True Substituting the point $(2,-2)$ into the equation makes it true. This point is on the graph of the equation.
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