Posted by Kimberly on Friday, March 1, 2013 at 9:07pm.
f'=+10(6x+6)/(3x^2+6x+6)^2
slope at x=-1
f'=0
y=mx+b=0+b
-10/3=b
y=-10/3 is the tangent line.
that wasnt right
is f(x)=-10/(3x^2+6x+6) or is
f(x)=-(10/3x^2 )+ 6x + 6 ?
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