Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.1 Quadratic Functions and Models - 3.1 Execises - Page 312: 4

Answer

-4

Work Step by Step

Recall that a parabola with vertex $(h, k)$ has a graph of the form $(y-k) = A(x-h)^2$ for some arbitrary constant A. Comparing with $f(x) = 2(x+4)^2 - 6$ (and with a little rearrangement), we can easily verify that our parabola's vertex is at $(-4, -6)$. Any line of symmetry of the parabola must, obviously, pass through its vertex. Since this is an upward-opening parabola (the leading coefficient of x > 0), and it is symmetric about the y-axis, the line of symmetry must be a line parallel to the y-axis that passes through $(-4, -6)$. This can only be $x = -4$!
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.