Calculate in what quadrant is the vertex of the graph of f=x^2-5x+6

The vertex of a parabola   y = ax^2+by+c
is

given by:  (xV,yV ) =  (-b/2a , 
-(b^2-4ac)/4a)....(1)

So the given parabola is y =
x^2-5x+6

a = 1, b = -5 and c =
6

Therefore substituting in
(1)

(xV , yV) ={ - -5/2*1 , ((-5)^2 -4*1+6))/4*1
}

= (5,  1/4) which is in 1st
quadrant.