math
posted by Kimberly .
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 = ?

math 
Knights
already answered by bobpursley

math 
Kimberly
he was wrong

math 
Reiny
Kimberly, as I pointed out to you in another post, it is very important to use brackets in this format to establish the correct order of operation, since we cannot "build" fractions here.
bobpursley interpreted your equation as
f(x) = 10/(3x^2 + 6 x + 6)
and his answer is correct according to that equation
the way you typed it ....
f'(x) = (20/3)x + 6
f'(1) = 20/3 + 6 =38/3
so the slope is 38/3
equation:
y + 10/3 = (38/3)(x+1)
times 3
3y + 10 = 38(x+1)
3y = 38x + 38  10
3y = 38x + 28
y = (38/3)x + 28/3
or
38x  3y = 28
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