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: 14

Answer

$(0,0)$ and $(1,-1)$ are on the graph.

Work Step by Step

In order to determine if a point is on the graph of the equation, we have to substitute into the equation: $y=x^3-2\sqrt{x}$ For the point of $(0,0)$ substitute $x=0$ and $y=0$ into the equation $0=0^3-2\sqrt{0}\\ 0=0$ The equation is satisfied, therefore the point is on the graph. For $(1,1)$: $x=1$ and $y=1$ $1=1^3-2\sqrt{1}\\ 1\ne-1$ The equation is not satisfied, therefore the point is not on the graph. For $(1,-1)$: $x=1$ $y=-1$ $-1=1^3-2\sqrt{1}\\ -1=-1$ The equation is satisfied, therefore the point is on the graph.
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