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March 4, 2015

March 4, 2015

Posted by **Lena** on Tuesday, February 24, 2009 at 8:11pm.

y=-3x^2 + 4x

This is how I solved it:

y=-3x^2 + 4x

y= -3 (x^2 + 4/3x)

y= -3(x^2+4/3x +4/9) + 4/3

y= -3 (x+2/3)^2 + 4/3

The answer sheet said that x = 2/3 but wouldn't it equal -2/3 because there is a -3 before the bracket?

- Find the minimum/maximum -
**bobpursley**, Tuesday, February 24, 2009 at 8:23pmy=-3x^2 + 4x

y= -3 (x^2 + 4/3x)You erred here, it should be

y=-3x^2 + 4x

y= -3 (x^2 - 4/3x )

- Find the minimum/maximum -
**Lena**, Tuesday, February 24, 2009 at 8:28pmoh ok

thanks so much :)

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