Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.3 Introduction to Graphing - R.3 Exercise Set: 41

Answer

$y=-x+3$

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, where \begin{array}{l}\require{cancel} x_1= 0 ,\\x_2= -1 ,\\y_1= 3 ,\\y_2= 4 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-3=\dfrac{3-4}{0-(-1)}(x-0) \\\\ y-3=\dfrac{3-4}{0+1}(x-0) \\\\ y-3=\dfrac{-1}{1}(x) \\\\ y-3=-x .\end{array} In $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-3=-x \\\\ y=-x+3 .\end{array}
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