Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 1 - Linear Functions - 1.1 Slopes and Equations of Lines - 1.1 Exercises - Page 13: 19


$y=-\frac{1}{3}x + \frac{10}{3}$.

Work Step by Step

RECALL: (i) The slope-intercept form of a line's equation is $y=mx+b$ where $m$ = slope and $b$ = y-intercept. (ii) The formula for slope is $m=\dfrac{y_2-y_1}{x_2-x_1}$. Solve for the slope using the formula above to have: $m=\dfrac{3-2}{1-4}=\dfrac{1}{-3}=-\dfrac{1}{3}$ Thus, the tentative equation of the line is $y=-\frac{1}{3}x + b$. Solve for the value of $b$ by substituting the x and y-coordinates of one point into the tentative equation to have: $y=-\frac{1}{3}x+b \\3 = -\frac{1}{3}(1)+b \\3=-\frac{1}{3}+b \\3+\frac{1}{3}=b \\\frac{10}{3}=b$ Therefore, the equation of the line in slope-intercept form is $y=-\frac{1}{3}x + \frac{10}{3}$.
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