College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.3 - Page 255: 31

Answer

$y=\frac{4}{3}x+2$

Work Step by Step

Use the slope formula $m=\frac{y^2-y^1}{x^2-x^1}$ and the point-slope form $y-y_1=m(x-x_1)$ to solve this problem. Passing through $(-3,-2)$ and $(3,6)$ $\frac{6+2}{3+3}=\frac{4}{3}$ $y+2=\frac{4}{3}(x+3)$ $y+2=\frac{4}{3}x+4$ $y=\frac{4}{3}x+2$
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