College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises: 27

Answer

$k(0)=3$ $k(2)=-5$ $k(-2)=3$ $k(\sqrt{2})=1-2\sqrt{2}$ $k(a+2)=-a^{2}-6a-5$ $k(-x)=-x^{2}+2x+3$ $k(x^{2})=-x^{4}-2x^{2}+3$

Work Step by Step

We are given: $k(x)=-x^{2}-2x+3$ We evaluate: $k(0)=-0^{2}-2(0)+3=0-0+3=3$ $k(2)=-2^{2}-2(2)+3=-4-4+3=-5$ $k(-2)=-(-2)^{2}-2(-2)+3=-4+4+3=3$ $k(\sqrt{2})=-(\sqrt{2})^{2}-2(\sqrt{2})+3=-2-2\sqrt{2}+3=1-2\sqrt{2}$ $k(a+2)=-(a+2)^{2}-2(a+2)+3=-(a^2+4a+4)-2a-4+3=-a^{2}-6a-5$ $k(-x)=-(-x)^{2}-2(-x)+3=-x^{2}+2x+3$ $k(x^{2})=-(x^{2})^{2}-2(x^{2})+3=-x^{4}-2x^{2}+3$
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