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: 12


$(0,1)$ and $(-1,0)$ are on the graph. $(1,2)$ is not on the graph.

Work Step by Step

For a point to be on the graph, it must satisfy the equation. Therefore we have to see for all three of the points if they satisfy the given equation: $y^3=x+1$. This can be don by substituting the values of $x$ and $y$ iof each point nto the equation. For the first point: $y^3=2^3=8$ $x+1=1+1=2$ $8\ne2$ Thus, the point is not on the graph. For the second point: $y^3=1^3=1$ $x+1=0+1=1$ $1=1$ Thus, the point is on the graph. For the third point: $y^3=0^3=0$ $x+1=-1+1=0$ $0=0$ Thus, the point is on the graph.
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