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: 75

Answer

$\color{blue}{f(x)=x^2-\frac{14}{3}x+\frac{8}{3}}$

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 $x=\frac{2}{3}$ and $x=4$. Using the rule mentioned in the recall part above, then the function is: $f(x)=(x-\frac{2}{3})(x-4) \\f(x)=x(x)+x(-4) - \frac{2}{3}(x) - \frac{2}{3}(-4) \\f(x)=x^2-4x-\frac{2}{3}x+\frac{8}{3} \\f(x)=x^2-\frac{12}{3}x-\frac{2}{3}x+\frac{8}{3} \\\color{blue}{f(x)=x^2-\frac{14}{3}x+\frac{8}{3}}$
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