Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.5 Equations of Lines and Linear Models - 2.5 Exercises - Page 242: 8



Work Step by Step

Solve for $y$ to obtain: $4x+3y=12 \\4x+3y-4x=12-4x \\3y=-4x+12 \\\frac{3y}{3}=\frac{-4x+12}{3} \\y=\frac{-4}{3}+4$ The slope of this line is $\frac{-4}{3}$. RECALL: The slope can be interpreted as the ratio of the rise (change in $y$) over run (change in $x$). The given equation has a slope of $\dfrac{-4}{3}$. This means that the rise is $-4$ units and the run is $3$ units. Notice for the graph in Option $B$, from the point $(0, 4)$ to the point $(3, 0)$, there is a rise of $-4$ unit and a run of $3$ units. Thus, its slope is $\dfrac{-4}{3}$. Therefore, the graph of the given equation is the one in Option $B$.
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