Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 3 - Section 3.1 - Reading Graphs and the Rectangular Coordinate System - Exercise Set - Page 202: 56


Point A: $y=\frac{1}{2}x+3$ $y=\frac{1}{2}(0)+3$ $y=3$ Point B: $y=\frac{1}{2}x+3$ $y=\frac{1}{2}(-4)+3$ $y=-2+3$ $y=1$ Point C: $y=\frac{1}{2}x+3$ $0=\frac{1}{2}x+3$ $-3=\frac{1}{2}x$ $x=-6$

Work Step by Step

To find the coordinates of Point A and B, substitute the given x-values into the given formula $y=\frac{1}{2}x+3$ to find the y-value for the corresponding x-values. To find the coordinates of Point C, substitute the given y-value into the given formula $y=\frac{1}{2}x+3$ to find the x-value for the corresponding x-value. When all three points are found, plot each point on the graph provided and draw a straight line, which represents (graphically) the given formula $y=\frac{1}{2}x+3$ .
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