Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 103: 51

Answer

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

Work Step by Step

$x-2y-1=0$ $;$ $(0,0)$ $,$ $(1,0)$ $,$ $(-1,-1)$ Substitute each point into the equation: $(0,0)$ $x-2y-1=0$ $0-2(0)-1=0$ $0-0-1=0$ $-1\ne0$ False Substituting the point $(0,0)$ did not make the left side of the equation equal to the right side. This point is not on the graph of the equation. $(1,0)$ $x-2y-1=0$ $1-2(0)-1=0$ $1-0-1=0$ $0=0$ True Substituting the point $(1,0)$ into the equation makes it true. This point is on the graph of the equation. $(-1,-1)$ $x-2y-1=0$ $-1-2(-1)-1=0$ $-1+2-1=0$ $0=0$ True Substituting the point $(1,0)$ into the equation makes it true. This point is on the graph of the equation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.