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

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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).**

To find the extreme value of y = x^2-8x+16.

Solution:

y = x^2-8x+16.

We see that the right is a complte square as x^2-8x+16 = (x-4)^2.

Therefore y = (x-4)^2.

Obviously (x-4)^2 , being a perfect square is always positive or > 0.

So (x-4)^2 > 0 , and (x-4) = 0 when x= 4.

Therefore y = 0 is the minimum , when x = 4.