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: 78



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 $x=15$ and $x=-12$. Using the rule mentioned in the recall part above, then the function is: $f(x)=c(x-15)[x-(-12)] \\f(x)=c(x-15)(x+12) \\f(x)=c[x(x)+x(12) - 15(x) - 15(12)] \\f(x)=c(x^2+12x-15x-180) \\f(x)=c(x^2-3x-180)$ With $f(3) = -45$, substitute $-45$ to $f(3)$ and $3$ to $x$ to obtain: $f(3) = c(3^2-3\cdot 3-180) \\-45 = c(9-9-180) \\-45=c(-180) \\\frac{-45}{-180}=c \\\frac{1}{4}=c$ Thus, the function is: $f(x) = \frac{1}{4}(x^2-3x-180) \\\color{blue}{f(x)=\frac{1}{4}x^2-\frac{3}{4}x-45}$
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