A function has a horizontal tangent line at the points where it's derivative is zero. (Since the derivative at a point is a slope of the tangent line at that point, if the derivative is zero, the tangent line is horizontal.)

The derivative of the given function is

`f'(x) = 3x^2 - 16x+5`

To find the zeros of the derivative, set up the equation

`f'(x) = 3x^2 - 16x + 5 = 0`

This quadratic equation can be solved by factoring:

`3x^2 - 16x + 5 = 3x^2 - 15x - x + 5 = 3x(x - 5) - (x - 5) `

`= (3x-1)(x-5)`

`(3x - 1)(x - 5)= 0`

3x - 1 = 0 and x- 5 = 0

`x = 1/3` and x = 5

Plug in these x-coordinates in `f(x)` in order to get the y-coordinates of the points:

`f(1/3) = (1/3)^3 - 8(1/3)^2 +5(1/3) +3=1/27-8/9 + 5/3 +3`

`=(1-24+45+81)/27=103/27`

`f(5) = 5^3 - 8*5^2 + 5*5 +3=125-200+25+3 =-47 `

**The coordinates of the points where the tangent line to f(x) is horizontal are**

**`(1/3, 103/27)` and (5, -47).**

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