Answer
$x=y^2-8y+12, 1\leq y\leq 6$
Work Step by Step
$x=t^2+4t=(2-y)^2+4(2-y)$
$x=y^2-8y+12$
Therefore,
$x=y^2-8y+12, 1\leq y\leq 6$
Also see attached graph.
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 10 - Review - Exercises - Page 690: 1
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