Intermediate Algebra (6th Edition)

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

Chapter 3 - Test - Page 198: 15



Work Step by Step

Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-PointForm, the equation of the line passing through $( 4,-2 )$ and $( 6,-3 )$ is \begin{array}{l} y-(-2)=\dfrac{-2-(-3)}{4-6}(x-4) \\\\ y+2=\dfrac{1}{-2}(x-4) \\\\ y+2=-\dfrac{1}{2}x+2 \\\\ y=-\dfrac{1}{2}x+2-2 \\\\ y=-\dfrac{1}{2}x .\end{array} In function notation, this is equivalent to $ f(x)=-\dfrac{1}{2}x .$
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