Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - Cumulative Standards Review - Multiple Choice - Page 190: 19

Answer

$f(-4) = 1$ $f(0) = 2$ $f(3) = \dfrac{11}{4}$

Work Step by Step

To find $f(-4)$, we plug $-4$ into the expression: $f(-4) = \frac{1}{4}(-4) + 2$ Multiply to simplify: $f(-4) = \frac{-4}{4} + 2$ Simplify the fraction: $f(-4) = -1 + 2$ Add to solve: $f(-4) = 1$ To find $f(0)$, we plug $0$ into the expression: $f(0) = \frac{1}{4}(0) + 2$ Multiply to simplify: $f(0) = 0 + 2$ Add to solve: $f(0) = 2$ To find $f(3)$, we plug in $3$ into the expression: $f(3) = \frac{1}{4}(3) + 2$ Multiply to simplify: $f(3) = \frac{3}{4} + 2$ Rewrite the constants as equivalent fractions: $f(3) = \frac{3}{4} + \frac{8}{4}$ Add to solve: $f(3) = \frac{11}{4}$
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