Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 3 - Quadratic Functions - 3.2 The Vertex of a Parabola - Exercises and Problems for Section 3.2 - Exercises and Problems - Page 127: 7

Answer

$ y=-\frac{1}{9}\cdot(x+6)^2+9 $

Work Step by Step

The vertex is the point $(h, k)=(-6,9)$, so the equation is in the form $$ y=a(x+6)^2+9 . $$ An $x$-intercept is -15, so $y=0$ when $x=-15$. This gives $$ \begin{aligned} a(-15+6)^2+9 & =0 \\ 81 a & =-9 \\ a & =-\frac{1}{9} \end{aligned} $$ Hence $$ y=-\frac{1}{9}\cdot(x+6)^2+9 $$
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