College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.5 - Equations of Lines and Linear Models - 2.5 Exercises - Page 227: 38

Answer

(a) 3 (b) -5 (c) $y=3x-5$

Work Step by Step

(a) To find the midpoint between points $(x_1,y_1)$ and $(x_2,y_2)$, we use the midpoint formula: $\displaystyle Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ We use the first two points in the table, $(-2,\ -11)$ and $(-1,\ -8)$: $midpoint=slope=\frac{-8-(-11)}{-1-(-2)}$ $=\frac{3}{1}$ $=3$ (b) To find the $y$-intercept, we look when $x=0$ in the table. This occurs at $y=-5$. Thus: y-intercept=$(0,-5)$ (c) A line in slope-intercept has the form: $y=mx+b$ ($m=slope$, $b=y-intercept$) We plug in $m=3$ and $b=-5$: $y=mx+b$ $y=3x-5$
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