Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - Review Exercises - Page 300: 17



Work Step by Step

Vertex: $(h,k)=(6,0)$ Standard form: $y=a(x-h)^2+k$ $y=a(x-6)^2+0$ $y=a(x-6)^2$ Now, use the point $(3,-9)$ to find $a$: $-9=a(3-6)^2+2$ $-9=a(9)$ $a=\frac{-9}{9}=-1$ $y=-(x-6)^2$
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