Calculus: Early Transcendentals (2nd Edition)

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

Chapter 1 - Functions - Review Exercises - Page 51: 3

Answer

$$\eqalign{ & {\text{a}}{\text{. }}y = \frac{5}{2}x - 8 \cr & {\text{b}}{\text{. }}y = \frac{3}{4}x + 3 \cr & {\text{c}}{\text{. }}y = \frac{1}{2}x - 2 \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{a}}.{\text{ The line passing through the points }}\left( {2, - 3} \right){\text{ and }}\left( {4,2} \right) \cr & {\text{The slope is }} \cr & m = \frac{{2 - \left( { - 3} \right)}}{{4 - 2}} = \frac{5}{2} \cr & {\text{The equation is }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = \frac{5}{2}\left( {x - 4} \right) \cr & y - 2 = \frac{5}{2}x - 10 \cr & y = \frac{5}{2}x - 8 \cr & \cr & {\bf{b}}.{\text{ The line with slope }}m = \frac{3}{4}{\text{ and }}x{\text{ - intercept }}\left( { - 4,0} \right) \cr & y = mx + b \cr & {\text{Calculating }}b{\text{ with the }}x{\text{ - intercept }}\left( { - 4,0} \right) \cr & 0 = \left( {\frac{3}{4}} \right)\left( { - 4} \right) + b \cr & 0 = - 3 + b \cr & b = 3 \cr & {\text{The equation is}} \cr & y = mx + b \cr & y = \frac{3}{4}x + 3 \cr & \cr & {\bf{c}}.{\text{ The line with intercepts }}\left( {4,0} \right){\text{ and }}\left( {0, - 2} \right) \cr & {\text{The slope is }} \cr & m = \frac{{ - 2}}{{ - 4}} = \frac{1}{2} \cr & {\text{The equation is }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 0 = \frac{1}{2}\left( {x - 4} \right) \cr & y = \frac{1}{2}x - 2 \cr & \cr & {\text{Graph}} \cr} $$
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