Algebra and Trigonometry 10th Edition

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

Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 249: 43

Answer

$y=-\frac{24}{49}(x+\frac{1}{4})^2+\frac{3}{2}$

Work Step by Step

Vertex: $(h,k)=(-\frac{1}{4},\frac{3}{2})$ Standard form: $y=a(x−h)^2+k$ $y=a[x-(-\frac{1}{4})]^2+\frac{3}{2}$ $y=a(x+\frac{1}{4})^2+\frac{3}{2}$ Now, use the point $(−2,0)$ to find $a$: $0=a(-2+\frac{1}{4})^2+\frac{3}{2}$ $-\frac{3}{2}=a(-\frac{7}{4})^2=a(\frac{49}{16})$ $a=\frac{-\frac{3}{2}}{\frac{49}{16}}=-\frac{3}{2}\frac{16}{49}=-\frac{24}{49}$ $y=-\frac{24}{49}(x+\frac{1}{4})^2+\frac{3}{2}$
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