Math Please Help!!
posted by Kimberly on .
Find the slope and the equation of the tangent line to the graph of the function f at the specified point.
f(x)=10/3x^2+6x+6; (1, 10/3)
slope ??
tangent line y = ?

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 wasn't right

is f(x)=10/(3x^2+6x+6) or is
f(x)=(10/3x^2 )+ 6x + 6 ?