Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 3 - Introduction to Graphing - Test: Chapter 3: 10

Answer

Refer to the graph below.
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Work Step by Step

Solve for $y$ by subtracting $1$ to both sides of the equation: $y+1-1=6-1 \\y=5$ This means that the given equation is equivalent to, and has the same graph as, $y=5$. RECALL: The graph of a line whose equation is of the form $y=k$ where $k$ is a real number is a horizontal line whose y-intercept is $(0, k)$. All points on the line have a y-coordinate of $k$. Thus, the graph of the given equation is a horizontal line whose y-intercept is $(0, 5)$. All the points on the line have a y-coordinate of $5$. The points $(-2, 5)$, $(0, 5)$, and $(2, 5)$ are all on the line. Plot these three points and connect them using a line to complete the graph. (Refer to the graph in the answer part above.)
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