Where is the vertex of the graph of f(x) = x^2 - 8x + 16 ?

To discover where the vertex of the parable f(x) = y is
located, we'll have to establish the quadrant or the side of x axis where the
coordinates of the vertex of the graph of y are located.

We
know that the coordinates of the parabola vertex
are:

 V(-b/2a;-delta/4a), where a,b,c are the coefficients
of the  function and delta=b^2 -4*a*c.

y=f(x)=x^2 - 8x +
16

We'll identify the
coefficients:

a=1, 2a=2,
4a=4

b=-8, c=16

delta=(-8)^2
-4*1*16

delta =64 - 64

delta =
0

V(-b/2a;-delta/4a)=V(-(-8)/2;-(0)/4)

V(4;0)

Because the x coordinate is
positive and y coordinate is 0, the vertex is located on the right side of x axis:
V(4;0).