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 - 3.3 Graphing and Intercepts - 3.3 Exercise Set - Page 180: 74

Answer

Refer to the graph below.

Work Step by Step

Divide both sides of the equation by $12$: $\dfrac{12y}{12}=\dfrac{-360}{12} \\y=-30$ This means that the given equation is equivalent to $y=-30$. Recall that the graph of an equation of the form $y=k$ is a horizontal line whose y-intercept is $(0, k)$. Each point on the line has a y-coordinate of $k$. Thus, the graph of $12y=-360$, which is the same as the graph of $y=-30$, is a horizontal line whose y-intercept is $(0, -30)$. Each point on the line has a y-coordinate of $-30$. This means that the following points are all on the line: $(8,-30), (0,-30), \text{ and } (-8,-30)$ Plot the three points listed above then connect them using a line to complete the graph. (Refer to the attached image in the answer part above for the graph.)
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