Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.1 - Graphs and Models - Exercises - Page 9: 74

Answer

One of the possible equations is $y= x^3-5x^2+\frac{1}{4}x+15.$

Work Step by Step

One of the way is to look for the polynomial of degree $3$ with three reals zeros $x_1=-3/2, x_2=4$ and $x_3=5/2$. According to the theorem of polynomial factorization we know that this polynomial has to be of the form of $$y=a(x-x_1)(x-x_2)(x-x_3)$$ where $a$ is nonzero real. Because we need only one function for simplicity just put $a=1$ which now yields $$y=(x+3/2)(x-4)(x-5/2)$$ or when expanded $$y= x^3-5x^2+\frac{1}{4}x+15.$$
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