Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Cumulative Review - Page 848: 3



Work Step by Step

Given $$y=|x+3|-8$$ Rewrite as: $y=|x-(-3)+(-8)$ Thus, we have $a=1\\k=-3\\h=-8$ Find the y-intercept: $y=|0+3|-8=-5$ Find another point: $x=1 \rightarrow y=-4\\x=2\rightarrow y=-3\\x=11\rightarrow y= 0$
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