Answer
See below
Work Step by Step
Given $$(x-3)^2=16y$$
Rewrite as: $$(x-3)^2=16(y-0)$$
Thus, we have $h=3\\k=4\\p=4 \gt 0$
The parabola opens upward.
The vertex is $(h,k)=(3,0)$
The focus is $(h,p+k)=(3,4)$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 12 Sequences and Series - Cumulative Review - Page 848: 23
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