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: 37


The function will have a minimum, and the value of the minimum will be $y=-2$.

Work Step by Step

The value of $A$ is positive, so the function will open upwards. Therefore, the function will have a minimum. The $x$-coordinate of the minimum will be $-\frac{B}{2A}$, if the function is in the form $Ax^{2}+Bx+C$. $A=\frac{3}{2}$, $B=6$, and $C=4$, so the $x$ coordinate will be: $=-6\div((2)\times(\frac{3}{2}))=-6\div(\frac{6}{2})=-\frac{6\times2}{6}=-\frac{12}{6}=-2$. Therefore, the value of the minimum is $y=-2$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.