Algebra 2 (1st Edition)

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

Chapter 4 Quadratic Functions and Factoring - Chapter Review - Page 322: 46


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Work Step by Step

The standard form of the equation is: $y=ax^2+bx+c$ Given three points: $(5,2)\\(0,2)\\(8,-6)$ Substitute: $5=a(2)^2+b(2)+c\\2=a(0)^2+b(0)+c\\-6=a(8)^2+b(8)+c$ We have the system: $4a+2b+c=5\\c=2\\64a+8b+c=-6$ Subtract the second equation from the first equation: $25a+5b=0\\ \rightarrow 5a+b=0$ (1) Subtract the first equation from the second equation: $64a+8b=-8\\ \rightarrow 8a+b=-1$ (2) Subtract equation (1) from equation (2): $3a=-1\\ \rightarrow a=-\frac{1}{3}$ Find $b$: $\frac{-25}{3})+5b+2=-2\\ \rightarrow b=\frac{5}{3}$ Hence, $a=\frac{-1}{3}\\b=\frac{5}{3}\\c=2$ Substitute back to the initial equation: $y=\frac{-1}{3}x^2+\frac{5}{3}x+2$
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