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: 77



Work Step by Step

RECALL: The zeros of the quadratic function $f(x) =c(x-a)(x-b)=0$ are $x=a$ and $x=b$. The given quadratic function has the zeros $x=4$ and $x=2$. Using the rule mentioned in the recall part above, then the function is: $f(x)=c(x-4)(x-2) \\f(x)=c[x(x)+x(-2) - 4(x) - 4(-2)] \\f(x)=c(x^2-2x-4x+8) \\f(x)=c(x^2-6x+8)$ With $f(0) = 40$, substitute $40$ to $f(0)$ and $0$ to $x$ to obtain: $f(0) = c(0^2-6\cdot 0+8) \\40 = c(0-0+8) \\40=c(8) \\\frac{40}{8}=c \\5=c$ Thus, the function is: $f(x) = 5(x^2-6x+8) \\\color{blue}{f(x)=5x^2-30x+40}$
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