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: 22


$y=\frac{21}{32}x +\frac{33}{16}$

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: $m=\dfrac{\frac{3}{4}-\frac{5}{2}}{-2-\frac{2}{3}}=\dfrac{-\frac{7}{4}}{-\frac{8}{3}}=\dfrac{-7}{4} \cdot \dfrac{-3}{8}=\dfrac{21}{32}$ Thus, the tentative equation of the line is $y=\frac{21}{32}x + b$. Solve for the value of $b$ by substituting the x and y-coordinates of one point into the tentative equation to have: $y=\frac{21}{32}x + b \\\frac{3}{4} = \frac{21}{32} \cdot (-2)+b \\\frac{3}{4}=-\frac{$21}{16}+b \\\frac{3}{4}+\frac{21}{16}=b \\\frac{33}{16}=b$ Therefore, the equation of the line in slope-intercept form is $y=\frac{21}{32}x +\frac{33}{16}$.
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