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: 26

Answer

$f(x)=2x+16$

Work Step by Step

Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, then the equation of the line passing through $ (-9,-2) \text{ and } (-3,10) ,$ is \begin{array}{l}\require{cancel} y-(-2)=\dfrac{-2-10}{-9-(-3)}(x-(-9)) \\\\ y+2=\dfrac{-2-10}{-9+3}(x+9) \\\\ y+2=\dfrac{-12}{-6}(x+9) \\\\ y+2=2(x+9) \\\\ y+2=2x+18 \\\\ y=2x+18-2 \\\\ y=2x+16 .\end{array} In function notation, this is equivalent to $ f(x)=2x+16 .$
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