Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.10 Write Quadratic Functions and Models - 4.10 Exercises - Skill Practice - Page 313: 39

Answer

See below

Work Step by Step

The standard form of the equation is: $y=ax^2+bx+c$ Given three points: $(-3,-2)\\(3,10)\\(6,-2)$ Substitute: $-2=a(-3)^2+b(-3)+c\\10=a(3)^2+b(3)+c\\-2=a(6)^2+b(6)+c$ We have the system: $9a-9b+c=-2\\9a+3b+c=10\\36a+6b+c=-2$ Subtract the second equation from the third equation: $27a+3b=-12\\ \rightarrow 9a+b=-4$ (1) Subtract the first equation from the second equation: $6b=12\\ \rightarrow b=2$ (2) Substitute $b$ in equation (1): $9a+2=-4\\ \rightarrow a=-\frac{2}{3}$ Find $c$: $9(\frac{-2}{3})+3(2)+c=-2\\ \rightarrow c=10$ Hence, $a=\frac{-2}{3}\\b=2\\c=10$ Substitute back to the initial equation: $y=\frac{-2}{3}x^2+2x+10$
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