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: 10

Answer

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

Work Step by Step

The points that are on the graph, must satisfy the equation. Therefore we have to see for all three of the points if they satisfy the given equation: $y=x^3-2\sqrt{x}$ For the first point: $x=0, y=0$ $y=x^3-2\sqrt{x}\\ 0=0^3-2\sqrt0\\ 0=0$ The point is on the graph. For the second point: $x=1, y=1$ $y=x^3-2\sqrt{x}\\ 1=1^2-2\sqrt1\\ 1\ne-1$ The point is not on the graph. For the third point: $x=1, y=-1$ $y=x^3-2\sqrt{x}\\ -1=1^3-2\sqrt1\\ -1=-1$ The point is on the graph.
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