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January 30, 2015

January 30, 2015

Posted by **mary** on Tuesday, January 17, 2012 at 10:40am.

- college algebra -
**Steve**, Tuesday, January 17, 2012 at 3:36pmfor any parabola, the min/max occurs where x = -b/2a

In your case, at x=6/6 = 1

domain for any polynomial is all reals

f(x) = 3x^2 - 6x - 7

= 3(x-1)^2 - 10

from this you can easily see that the min at x=1 is -10

so the range is all reals >= 10

- college algebra -
**M**, Sunday, January 22, 2012 at 11:06pmBy completing square,

f(x) = 3[x^2 -2x - 7/3]

= 3[x^2 - 2(x)(1) + (1)^2 - (1)^1 - 7/3]

= 3[(x - 1)^2 - 1 - 7/3]

= 3(x - 1)^2 - 3 - 7

= 3(x - 1)^2 - 10

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