Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 1.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - 1.2 Assess Your Understanding - Page 17: 15

Answer

Only $(0,3)$ is on the graph of $y^{2}=x^{2}+9$.

Work Step by Step

a) If $(0,3)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=0$ and $y=3$, we get- $(3)^{2}=(0)^{2}+9$ $9=9$ Since LHS=RHS therefore $(0,3)$ lies on the graph. b) If $(3,0)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=3$ and $y=0$, we get- $(0)^{2}=(3)^{2}+9$ $0\ne18$ Since LHS$\ne$RHS therefore $(3,0)$ does not lie on the graph. c) If $(-3,0)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=-3$ and $y=0$, we get- $(0)^{2}=(-3)^{2}+9$ $0\ne18$ Since LHS$\ne$RHS therefore $(-3,0)$ does not lie on the graph.
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