Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.3 - Systems of Linear Equations and Problem Solving - Exercise Set - Page 234: 54

Answer

$a=1$ $b=-2$ $c=3$

Work Step by Step

Given Equation, $y=ax^{2}+bx+c$ Given ordered pair solutions $(1,2), (2,3),(-1,6)$ Substituting each ordered pair solution into the equation, $y=ax^{2}+bx+c$ $2=a(1)^{2}+b(1)+c$ $a+b+c=2$ Equation $(1)$ $y=ax^{2}+bx+c$ $3=a(2)^{2}+b(2)+c$ $4a+2b+c=3$ Equation $(2)$ $y=ax^{2}+bx+c$ $6=a(-1)^{2}+b(-1)+c$ $a-b+c=6$ Equation $(3)$ Adding Equation $(1)$ and Equation $(3)$ $a+b+c+a-b+c=2+6$ $2a+2c=8$ $a+c=4$ Equation $(4)$ Substituting Equation $(4)$ in Equation $(1)$ $a+b+c=2$ $b+4=2$ $b=-2$ Subtracting Equation $(3)$ From Equation $(2)$ $4a+2b+c-(a-b+c)=3-6$ $4a+2b+c-a+b-c=-3$ $3a+3b=-3$ $a+b=-1$ Substituting $b$ value $a-2=-1$ $a=-1+2$ $a=1$ Substituting $a$ and $b$ values in Equation $(4)$ $a+c=4$ $1+c=4$ $c=3$
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