Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 1.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - 1.2 Assess Your Understanding - Page 18: 88

Answer

$y=\frac{1}{8}x^2 -\frac{3}{4}x +1$

Work Step by Step

$y$ = a$x^2$ + b$x$ + c Substitute from the equation $(2,0), (4,0),$ and $(0,1)$ $0 = 2^2$a + $2$b +c $0 = 4^2$a + $4$b +c $1 = 0^2$a + $0$b +c $0 = 4$a + $2$b +c $0 = 16$a + $4$b +c $1 = $ c By simultaneous equation calculator a $=\frac{1}{8}$ b $=-\frac{3}{4}$ c $=1$ Therefore, $y=\frac{1}{8}x^2 -\frac{3}{4}x +1$
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