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October 2, 2014

October 2, 2014

Posted by **ANONYMOUS....** on Tuesday, March 3, 2009 at 2:50pm.

f(x)=2x^2-4x

- ALGEBRA -
**Reiny**, Tuesday, March 3, 2009 at 3:28pmif 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

- ALGEBRA -
**Damon**, Tuesday, March 3, 2009 at 3:30pmy = 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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