Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 82: 162

Answer

The graph of the equation is a straight line with points $\left(-3,-7\right),\ \left(-2,-5\right),\left(-1,-3\right),\left(0,-1\right),\left(1,1\right),\left(2,3\right),\left(3,5\right)$ obtained by substituting the given values of $x$ to the equation.

Work Step by Step

$$y = 2x - 1$$ If $x=-3$ $y = 2(-3) - 1$ $y = -6 - 1$ $y = -7$ If $x=-2$ $y = 2(-2) - 1$ $y = -4 - 1$ $y = -5$ If $x=-1$ $y = 2(-1) - 1$ $y = -2 - 1$ $y = -3$ If $x=0$ $y = 2(0) - 1$ $y = 0 - 1$ $y = -1$ If $x=1$ $y = 2(1) - 1$ $y = 2- 1$ $y = 1$ If $x=2$ $y = 2(2) - 1$ $y = 4 - 1$ $y = 3$ If $x=3$ $y = 2(3) - 1$ $y = 6 - 1$ $y = 5$ Thus, the graph will have the points $\left(-3,-7\right),\ \left(-2,-5\right),\left(-1,-3\right),\left(0,-1\right),\left(1,1\right),\left(2,3\right),\left(3,5\right)$.
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