Answer
See below
Work Step by Step
Given: $y=\frac{1}{4}(x-3)^2+2$
We can see that $a=\frac{1}{4}\\h=3\\k=2$
The vertex is: $(h,k)=(3,2)$
Find some points
$x=0\rightarrow y=4.25\\x=6 \rightarrow y=4.25$

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