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: 79



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 $x=\frac{1}{4}$ and $x=-2$. Using the rule mentioned in the recall part above, then the function is: $f(x)=c(x-\frac{1}{4})[x-(-2)] \\f(x)=c(x-\frac{1}{4})(x+2) \\f(x)=c[x(x)+x(2) - \frac{1}{4}(x) - \frac{1}{4}(2)] \\f(x)=c(x^2+2x-\frac{1}{4}x-\frac{1}{2}) \\f(x)=c(x^2+\frac{7}{4}x-\frac{1}{2})$ With $f(2) = -21$, substitute $-21$ to $f(2)$ and $2$ to $x$ to obtain: $f(2) = c(2^2+\frac{7}{4}\cdot 2-\frac{1}{2}) \\-21 = c(4+\frac{7}{2}-\frac{1}{2}) \\-21 = c(\frac{8}{2}+\frac{6}{2}) \\-21=c(\frac{14}{2}) \\-21=c(7) \\\frac{-21}{7}=c \\-3=c$ Thus, the function is: $f(x) = -3(x^2+\frac{7}{4}x-\frac{1}{2}) \\\color{blue}{f(x)=-3x^2-\frac{21}{4}x+\frac{3}{2}}$
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