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: 41

Answer

(a) y-intercept=$(0,2)$ x-intercept=$(6,0)$ slope=$-1/3$ (b) $y=-1/3x+2$

Work Step by Step

(a) We find the $x$-intercept by looking at the graph for the point at which $y=0$. Similarly, we find the $y$-intercept by looking for the point at which $x=0$. y-intercept=$(0,2)$ x-intercept=$(6,0)$ To calculate the slope between points $(x_1,y_1)$ and $(x_2,y_2)$, we use the formula: $slope=m=\frac{y_2-y_1}{x_2-x_1}$ We plug in the $x$ and $y$ intercept points: slope=$\frac{0-2}{6-0}=-1/3$ (b) A line in slope-intercept has the form: $y=mx+b$ ($m=slope$, $b=y-intercept$) We plug in $m$ and $b$ from part (a): $y=-1/3x+2$
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