Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 72

Answer

$\color{blue}{f(x)=x^2+14x+45}$

Work Step by Step

RECALL: The zeros of the quadratic function $f(x) =(x+a)(x+b)=0$ are $x=-a$ and $x=-b$. The given quadratic function has the zeros $x=-5$ and $x=-9$. Using the rule mentioned in the recall part above, then the function is: $f(x)=(x+5)(x+9) \\f(x)=x^2+x(9) + 5(x) + 5(9) \\f(x)=x^2+9x+5x+45 \\\color{blue}{f(x)=x^2+14x+45}$
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