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: 14


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 $x$. $f(x) = \frac{2}{3}x+2 \\0=\frac{2}{3}x+2 \\0-2=\frac{2}{3}x+2-2 \\-2=\frac{2}{3}x \\-2(\frac{3}{2})=2(\frac{1}{2}x)(\frac{3}{2}) \\-3=x$ The x-intercept is $(-3, 0)$. Find the y-intercept by setting $x=0$ then solving for $y$. $f(x)=\frac{2}{3}x+2 \\f(x) = \frac{2}{3}(0)+2 \\f(x)=0+2 \\f(x)=2$ The y-intercept is $(0, 2)$. Plot the intercepts and connect them using a line. The point $(3, 4)$ can be used as a check point. (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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