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: 15

Answer

point-slope form: $y+3=-3(x+2)$ slope-intercept form: $y=-3x-9$

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 $-3$ and passes through the point $(-2, -3)$. Substitute these values into the point-slope form above to obtain: $y-(-3)=-3[x-(-2)] \\y+3 = -3(x+2)$ (b) slope-intercept form Substitute the slope $-3$ to $m$ to obtain the tentative equation: $y=-3x+b$ The line passes through $(-2, -3)$. 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: $y=-3x+b \\-3 = -3(-2) + b \\-3 = 6 + b \\-3-6 = b \\-9= b$ Thus, the equation of the line is $y=-3x-9$.
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