Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - 3.3 Quadratic Functions and Their Properties - 3.3 Assess Your Understanding - Page 145: 14



Work Step by Step

If $f(x)=ax^2+bx+c$, then if $a\gt0$, the graph opens up, hence it has a maximum, if $a\lt0$, the graph opens down, hence it has a minimum. The minimum value of the graph is at $x=-\frac{b}{2a}$. Let's compare $f(x)=-x^2-1$ to $f(x)=ax^2+bx+c$. We can see that $a=-1$, $b=0$, $c=-1$. $a\lt0$, hence the graph opens down, hence it's vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{0}{2\cdot(1)}=0.$ Hence the maximum value is $f(0)=-(0)^2-1=-1.$ Thus, the vertex is at $(0,-1)$. The vertex is at $(0,-1)$ and the graph opens down. Hence it is graph $E$.
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.