Answer
Please see the graph.
Work Step by Step
$y=x^2+4$
vertex:
$x=-b/2a$
$x=-0/2*1$
$x=0$ (axis of symmetry)
$y=x^2+4$
$y=0^2+4$
$y=0+4=4$
$(0,4)$
One other point on the line:
$x=1$
$y=x^2+4$
$y=1^2+4$
$y=1+4=5$
$(1,5)$
Algebra 1
by Hall, Prentice
Published by
Prentice Hall
ISBN 10:
0133500403
ISBN 13:
978-0-13350-040-0
Chapter 10 - Radical Expressions and Equations - Get Ready - Page 597: 13
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