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: 10


Point-slope form" $y+4=-(x+\frac{1}{4})$ Function notation of the slope-intercept form: $f(x)=-x-\frac{17}{4}$

Work Step by Step

RECALL: (i) The point-slope form of a line's equation is: $y-y_1=m(x-x_1)$ where m= slope and $(x_1, y_1)$ is a point on the line. (ii) The function notation of the slope-intercept form of a line's equation is: $f(x) = mx + b$ where m= slope and b = y-intercept The given line has $m=-1$ and passes through the point $(-\frac{1}{4}, -4)$. This means that the point-slope form of the line's equation is: $y-(-4) = -1[x-(-\frac{1}{4})] \\y+4=-(x+\frac{1}{4})$ Convert the equation to slope-intercept form by isolating $y$ to obtain: $y + 4 =-(x+\frac{1}{4}) \\y+4=-1\cdot x + (-1)\cdot \frac{1}{4} \\y+4=-x+(-\frac{1}{4}) \\y+4=-x-\frac{1}{4} \\y+4-4=-x-\frac{1}{4}-4 \\y=-x-\frac{1}{4} - \frac{16}{4} \\y=-x-\frac{17}{4}$ In function notation, the slope-intercept form of the equation is: $f(x) = -x-\frac{17}{4}$
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