Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-2 Quadratic Functions\ - Practice and Problem-Solving Exercises - Page 544: 15


Axis of symmetry: $x= -\frac{1}{2}$ Vertex: $(-\frac{1}{2},-\frac{13}{2})$

Work Step by Step

$y = 6x^{2} + 6x - 5$ The standard form for a quadratic equation is $y = ax^{2} + bx + c$ So a= 6, b= 6, and c= -5 Axis of symmetry: The formula for axis of symmetry is $x= \frac{-b}{2a}$ $x= \frac{-(6)}{2(6)}$ $x= \frac{-6}{12}$ $x= -\frac{1}{2}$ Vertex: Plug in the x value of the axis of symmetry to find the y value of the vertex. $y = 6x^{2} + 6x - 5$ $y = 6(-\frac{1}{2})^{2} + 6(-\frac{1}{2}) - 5$ $y= -\frac{13}{2}$ The vertex is $(-\frac{1}{2},-\frac{13}{2})$
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.