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