Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 3 - Section 3.5 - Equations of Lines - Exercise Set - Page 173: 28

Answer

$f(x)=\frac{5}{4}x-13$

Work Step by Step

We are given the points $(8,-3)$ and $(4,-8)$. We can use the slope formula to find the slope of a line going through these points. $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-8-(-3)}{4-8}=\frac{-5}{-4}=\frac{5}{4}$ Next, we can use this slope and one of the given points to write the line in the point slope form $y-y_{1}=m(x-x_{1})$. $f(x)=y+3=\frac{5}{4}(x-8)=\frac{5}{4}x-10$ Subtract 3 from both sides. $f(x)=\frac{5}{4}x-13$
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