Posted by ANONYMOUS.... on Tuesday, March 3, 2009 at 2:50pm.
if you know Calculus, then
f'(x) = 4x - 4
= 0 for a max/min
4(x-1) = 0
x = 1
then f(1) = 2-4 = -2
so (1,-2) is a min (since the parabola opens upwards)
if you don't know Calculus, use "completing the square" to find the vertex
f(x) = 2[x^2 - 2x + 1 - 1 ]
= 2[(x-1)^2 - 1]
= 2(x-1)^2 - 2
so the vertex is (1,-2) as above
or
you could use the formula:
the x of the vertex is -b/(2a)
= -(-4)/2(2)
= -1
etc
y = 2 (x^2-2x)
dy/dx = 2 (2x-2) = extreme at x = 1
d^2y/dx^2 = 4 so that extreme is a minimum
at x = 1
f(1) = -2
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