Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX C - Graphs of Second-Degree Equations - C Exercises - Page A 23: 30

Answer

$x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$

Work Step by Step

$y^2-2x+6y+5 = 0$ $2x = y^2+6y+5$ $2x = y^2+6y+9-9+5$ $2x = (y+3)^2-4$ $x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$
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