Find the vertex of the parabola y=x^2-6x?

We'll recall what are the coordinates of the vertex of the
parabola:

V(-b/2a ;
-delta/4a)

a,b,c are the coefficients of the
quadratic:

y = ax^2 + bx +
c

Comparing, we'll identify the
coefficients:

a = 1, b = -6 , c =
0

We'll calculate the x coordinate of the
vertex:

xV = -b/2a

xV =
-(-6)/2

xV = 3

We'll calculate
the y coordinate of the vertex:

yV =
-delta/4a

delta = b^2 -
4ac

delta = 36 - 4*1*0

delta =
36

yV = -36/4

yV =
-9

The requested coordinates of the vertex of
the given parabola are: V(3 , -9).