To determine the point of maxima for the curve y = x^2 - 8x + 16, we need to find the derivative of y and equate it to zero.

y = x^2 - 8x + 16

=> y' = 2x - 8

2x - 8 = 0

=> x = 8/2

=> x = 4

For x = 4, y = 4^2 - 8*4 + 16

=> y = 16 - 32 + 16 = 0

But we see that the point (4, 0) is an extreme point where the function has the minimum value.

**It is not possible to determine the point where the function has the maximum value.**

To determine the maximum point of the curve y = x^2 - 8x + 16

y = x^2 - 8x + 16

y = (x-4)^2 is a perfect square which is always greater than or equal to zero. Y = 0 when x= 4.

y > = 0 for all x.

Therefore y is minimum when x= 4.

As x increses positively y also increases positively.

Lt x-->infinity y = Lt x-> infinity (x-4)^2 is also infinity.

Lt x-> -infinity y = Lt x-> -infinity (x-4)^2 = -infinity.

So y is unbounded and has no maximum.

The maximum point of these quadratic function is represented by the vertex of the function.

The graph of the quadratic function is a parabola and 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(-b/2a;-delta/4a)=V(4;0) **

**We notice that the x coordinate is positive and y coordinate is 0, so the vertex of parabola is located on the right side of x axis: V(4;0).**