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 - Guided Practice for Examples 1, 2, and 3 - Page 310: 3

Answer

$f(x)=0.25(x+2)(x-5)$.

Work Step by Step

If the x-intercepts of a graph are $a,b$, then the function has the form $f(x)=k(x-a)(x-b)$. The x-intercepts are -2, 5, hence the quadratic function becomes $f(x)=k(x+2)(x-5)$. The point (6,2) is on the graph, hence if we plug in the values we get $2=k(8)(1)$ and k=0.25. $f(x)=0.25(x+2)(x-5)$.
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