Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Section 2.5 - The Point-Slope Form of the Equation of a Line - Exercise Set - Page 163: 21


$ y-9=-\displaystyle \frac{11}{3}(x+1) \qquad$ ... point-slope form $y=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... slope-intercept form $f(x)=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... function notation

Work Step by Step

$(x_{1},y_{1})=(1,9) \; (x_{2},y_{2})=(4,-2)$ $m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-9}{4-1}=\frac{-11}{3}=-\frac{11}{3}$. So with $m=-\displaystyle \frac{11}{3}$ and $(x_{1},y_{1})=(1,9)$, we write the point-slope form $y-y_{1}=m(x-x_{1})$ $ y-9=-\displaystyle \frac{11}{3}(x-1) \qquad$ ... point-slope form Simplify to slope-intercept form, $ y=mx+b$ ... distribute $ y-9=-\displaystyle \frac{11}{3}x+\frac{11}{3} \qquad$ ...add $9$ $ y=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... is the slope-intercept form For function notation, replace $y$ with $f(x)$.
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