Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 29

Answer

Parabola Focus: $( 1,-\frac{11}{6})$ Vertex: $(1,-2)$

Work Step by Step

$3x^2-6x-2y=1$ $3(x^2-2x)=2y$ $3(x-1)^2-4=2y$ $y=\frac{3}{2}(x-1)^2-2$, the equation of an upwards opening parabola. $(h,k+p)=(1,-2+1/6)=( 1,-\frac{11}{6})$ Focus: $( 1,-\frac{11}{6})$ Vertex: $(1,-2)$
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