Answer
Vertex, (0,3)
Axis of symmetry, x=0
Work Step by Step
Given the function
$y= \frac{1}{4}x^{2} + 3$
We need to find the vertex and axis of symmetry
$y=a(k(x-d))^{2} + c$
Vertex:
The vertex for the base function $y= x^{2}$ is (0,0).
Since the d value is zero, the graph does not shift left or right thus the x value of the vertex is 0.
Since the c value is 3, the graph moves three units up thus the y value of the vertex is 3.
Therefore the vertex is (0,3)
Axis of symmetry:
The Axis of symmetry of a $y= x^{2}$ graph is x=0. Since the d value is zero, the graph does not shift left or right thus the axis of symmetry is still x=0.
