math
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 = ?

already answered by bobpursley

he was wrong

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