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: $x=1$ $y=x^2+4$ $y=1^2+4$ $y=1+4=5$ $(1,5)$
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.