#### Answer

$-10-4+0$
$=-14$

#### Work Step by Step

$\sum_{i=1}^{3}(3x_i-x_i^2)$
$=(3x_1-x_1^2)+(3x_2-x_2^2)+(3x_3-x_3^2)$
$=(3*(-2)-(-2)^2)+(3*(-1)-(-1)^2)+(3*0-0^2)$
$=-10-4+0$
$=\boxed{-14}$

Published by
Pearson

ISBN 10:
013421742X

ISBN 13:
978-0-13421-742-0

$-10-4+0$
$=-14$

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