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.
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 = ?
3 answers
that wasn't right
is f(x)=-10/(3x^2+6x+6) or is
f(x)=-(10/3x^2 )+ 6x + 6 ?
f(x)=-(10/3x^2 )+ 6x + 6 ?
2 answers