College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.1 - Polynomial Functions and Models - 5.1 Assess Your Understanding - Page 339: 73

Answer

$f(x)=ax(x-1)(x-2), \text{ where }a>0$

Work Step by Step

Thanks to the real zeros we see on the graph, we can find the function through its factored form: $f(x)=a(x-0)(x-1)(x-2)$ That's because when any (x-r) equals zero, f(x) equals zero. The only thing left is to find 'a'. Nonetheless, since there are no non-x-intercept points given to solve the equation for 'a', we can leave the answer as: $f(x)=ax(x-1)(x-2), \text{ where }a>0$ The reason 'a' must be a positive number is that the graph resembles $y=x^3$. If 'a' were to be negative, the graph will look flipped like $y=-(x^3)$.
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