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: 76



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 $x=6$ and $x=-1$. Using the rule mentioned in the recall part above, then the function is: $f(x)=c(x-6)[x-(-1)) \\f(x)=c(x-6)(x+1) \\f(x)=c[x(x)+x(1) - 6(x) - 6(1)] \\f(x)=c(x^2+x-6x-6) \\f(x)=c(x^2-5x-6)$ With $f(2) = -36$, substitute $-36$ to $f(x)$ and $2$ to $x$ to obtain: $f(2) = c(2^2-5\cdot 2-6) \\-36 = c(4-10-6) \\-36=c(-12) \\\frac{-36}{-12}=c \\3=c$ Thus, the function is: $f(x) = 3(x^2-5x-6) \\\color{blue}{f(x)=3x^2-15x-18}$
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