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

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You may evaluate the maximum of the given function using derivative of the function, such that:

`f'(x) = 0`

Evaluating `f'(x) ` yields:

`f'(x) = (x^2 - 8x + 16)' => f'(x) = 2x - 8`

`f'(x) = 0 => 2x - 8 = 0 => 2x = 8 => x = 4`

**Hence, the function `f(x) = x^2 - 8x + 16` reaches its maximum at **`x = 4.`

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).**

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