Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 41

Answer

$$y = - \frac{1}{2}x + 12$$

Work Step by Step

$$\eqalign{ & y{\text{ - intercept }}\underbrace {\left( {0,12} \right)}_{\left( {0,b} \right)} \cr & x + 2y = 8 \to \underbrace {y = - \frac{1}{2}x + 4}_{y = mx + b} \cr & {\text{The line is parallel to the line }}x + 2y = 8,{\text{so they have the same}} \cr & {\text{slope.}}{\text{. Then }} \cr & m = - \frac{1}{2}{\text{ and }}y{\text{ - intercept }}\underbrace {\left( {0,12} \right)}_{\left( {0,b} \right)} \cr & {\text{Use the slope - intercept form of the equation of a line}} \cr & y = mx + b,{\text{ where }}m = slope,{\text{ and }}\left( {0,b} \right){\text{ is }}y{\text{ - intercept}} \cr & y = - \frac{1}{2}x + 12 \cr & \cr & {\text{Graph}} \cr} $$
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