Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.4 - Linear Functions and Slope - Exercise Set: 19

Answer

point-slope form: $y+2=-(x+\frac{1}{2})$ slope-intercept form: $y=-x-2.5$

Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where m = slope and b = y-intercept (2) The point-slope form of a line's equation is: $y-y_1=m(x-x_1)$ (a) point-slope form The given line has a slope of $-1$ and passes through the point $(-\frac{1}{2}, -2)$. Substitute these values into the point-slope form above to obtain: $y-(-2)=-1[x-(-\frac{1}{2})] \\y+2 = -(x+\frac{1}{2})$ (b) slope-intercept form Substitute the slope $-1$ to $m$ to obtain the tentative equation: $y=-x+b$ The line passes through $(-\frac{1}{2}, -2)$. This means that the coordinates of this point satisfy the equation of the line. Substitute the x and y-coordinates of this point into the tentative equation to obtain: $y=-x+b \\-2 = -(-\frac{1}{2}) + b \\-2 = \frac{1}{2} + b \\-2-\frac{1}{2}= b \\-2.5 = b$ Thus, the equation of the line is $y=-x-2.5$.
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