Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1 - Linear Functions - 1.6 Finding Linear Models - 1.6 Exercises - Page 93: 41

Answer

$y=6x-20$

Work Step by Step

The plot of the points corresponding to the given data is shown below. It also shows that the given data can be modeled by the line that passes through $(4,4)$ and $(12,52)$ since this line would pass closely to all the given points. The equation of the line passing through the two points above takes the form of $$ y=mx+b, \tag{1}$$where $m$ is the slope and $b$ is the $y$-intercept. Using the formula for finding the slope, $m$, of the line passing through two points, $(x_1,y_1)$ and $(x_2,y_2)$, which is given by $m=\frac{y_1-y_2}{x_1-x_2}$, then $$\begin{aligned} m&=\frac{y_1-y_2}{x_1-x_2} \\&= \frac{4-52}{4-12} \\&= \frac{-48}{-8} \\&= 6 .\end{aligned} $$With $m=6$, then Eqn. $(1)$ becomes $$ y=6x+b .\tag{2}$$Since the line passes through the point $(4,4)$, substitute $x=4$ and $y=4$ in Eqn. $(2)$ to solve for $b$. That is, $$\begin{aligned} y&=6x+b \\ 4&=6(4)+b \\ 4&=24+b \\ 4-24&=b \\ -20&=b .\end{aligned} $$With $b=-20$, then Eqn. $(2)$ becomes $$ y=6x-20 .$$Hence, the equation of the line that models the given data set is $y=6x-20$.
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