## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 2 Linear Equations and Functions - 2.1 Represent Relations and Functions - 2.1 Exercises - Skill Practice - Page 78: 38

-208

#### Work Step by Step

For starters, the function is not linear, for the slope is not a constant value due to the $x^3$ and $x^2$ terms. We now evaluate $g(-5)$: $$g(-5) = (-5)^3 -2 (-5)^2 +5(-5) - 8 \\ g(-5) = -125 -50 -25 - 8 \\ g(-5) = -208$$

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