What is y=(x-3)^2+4 in standard form

Standard form of a parabola:y=(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. Parabola opens upward if coefficient of (x-h)^2 term is positive and downward if the coefficient is negatve.
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Given parabola opens downward.(by inspection)
Vertex: (3,4) (by inspection)
Axis of symmetry: x=3 (by inspection)
solving for y-intercept

let x=0
y=-9+4=-5
y-intercept=5
solving for x-intercepts
let y=0
(x-3)^2=4
x-3=+-sqrt(4)
x=3+-2
x=1
x=5 

(3,4) is the answer to your question. focus is (3, 17/4) axis of symmetry is x=3 directrix is y= 15/4. hope this helps.

Um what? rephrase that ?