Answer
$c-\frac{b^2}{4a}$
Work Step by Step
For $a\gt0$, the least value of $f(x)=ax^2+bx+c$ is the y-coordinate of the vertex.
We have $x=-\frac{b}{2a}$ and $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c=c-\frac{b^2}{4a}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 3 - Polynomial and Rational Functions - Chapter 3 Test Prep - Review Exercises - Page 397: 8
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
