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: 16

Answer

$(0,1)$ and $(-1, 0)$ 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: $y^3=x+1$ For the point of $(1,2)$ substitute $x=1$ and $y=2$ into the equation $2^3=1+1\\ 8\ne2$ The equation is not satisfied, therefore the point is not on the graph. For $(0,1)$: $x=0$ and $y=1$ $1^3=0+1\\ 1=1$ The equation is satisfied, therefore the point is on the graph. For $(-1,0)$: $x=-1$ $y=0$ $0^3=-1+1\\ 0=0$ The equation is satisfied, therefore the point is on the graph.
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