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
delta =64 - 64
delta = 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.
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.