Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.2 Representing Functions - 1.2 Exercises - Page 24: 56

Answer

(2,6) and (-1,3)

Work Step by Step

$y=x^{2}+2$ $y=x+4$ $x^{2}+2=x+4$ $x^{2}-x-2=0$ $(x-2)(x+1)=0$ $x=2,-1$ $y=2+4=6$ $y=-1+4= 3$ (2,6) and (-1,3)
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