Algebra 2 (1st Edition)

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

Chapter 4 Quadratic Functions and Factoring - 4.1 Quadratic Functions in Standard Form - 4.1 Exercises - Skill Practice - Page 241: 36


The function will have a maximum, and the value of the maximum is $y=9$.

Work Step by Step

The function's $A$-value is negative, so it will open downwards. Therefore, the function will have a maximum. In order to figure out the value of the $y$-coordinate of the maximum, we should take the value of the $x$-coordinate (which is $-\frac{B}{2A}$) and plug it in for the $x$-values in the function's equation. This would result in $y=-3\times(-\frac{B}{2A})^{2}+(18)\times(-\frac{B}{2A}-5)$. $A=-3$, and $B=18$. Substituting those values into the equation results in: $y=-3\times(-\frac{18}{2\times-3})^{2}+(18)\times(-\frac{18}{2\times-3}-5)$ $y=-3\times(3)^{2}+(18)\times(3-5)$ $y=-3\times(9)+18\times(-2)=-27+36=9$ Therefore, the value of the maximum is $y=9$.
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