What is the extreme of y=x^2-8x+16?

To extreme of a function, whose expression is a quadratic,
is the vertex of the parable f(x) = y.

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)

Since the coefficient of x^2
is positive, the extreme point is a minimum
point.

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).