Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 13



Work Step by Step

Standard form of a horizontal parabola is $x=a(y-k)^2+h$ Here, vertex:$(h,k)$ and $4a$ is latus rectum. Since we are given $x=2(y-6)^2$ It can be written as: $x=2(y-6)^2+0$ Compare the given equation with the standard form. Thus, vertex: $(0,6)$
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