Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.4 Linear Functions - 2.4 Exercises - Page 228: 16


domain: $(-\infty, +\infty)$ range: $(-\infty, +\infty)$ Refer to the graph below.

Work Step by Step

Find the x-intercept by setting $f(x)$ or $y$ to zero, then solve for $x$. $f(x) = -2x \\0=-2x \\\frac{0}{-2}=\frac{-2x}{-2} \\0=x$ The x-intercept is $(0, 0)$. Find the y-intercept by setting $x=0$ then solving for $y$. $f(x)=-2x \\f(x) = -2(0) \\f(x)=0$ The y-intercept is $(0, 0)$. Since the x and y-intercepts are the same, one more point is needed to graph the linear function. This can be found by assigning any value to $x$ then solving for $y$: If $x=1$, $f(1) = -2(1) \\f(1) = -2$ Thus, the point $(1, -2)$ is also on the line. Plot $(0, 0)$ and $(1, -2)$ and connect the points using a line to complete the graph. (Refer to the graph in the answer part above.) The graph covers all x-values and all y-values. Thus, the given function has: domain: $(-\infty, +\infty)$ range: $(-\infty, +\infty)$
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