# Find all those points where the tangent line to the curve y=x^3-x^2+1 is horizontal? Show your work

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You need to remember that when the tangent line to a curve becomes horizontal, its slopes is equal to zero.

Using the equation that relates the derivative and the slope at the tangency point, yields:

`f'(x) = m`

Since `m = 0` , yields:

`f'(x) = 0`

Differentiating the function with respect to `x` , yields:

`f'(x) = (x^3-x^2+1)' => f'(x) = 3x^2 - 2x`

You need t solve for x the following equation, such that:

`3x^2 - 2x = 0`

Factoring out x yields:

`x(3x - 2) = 0 `

Using the zero product rule yields:

`x = 0`

`3x - 2 = 0 => 3x = 2 => x = 2/3`

Since the problem requires for you to find the points where the tangent line to the curve `y=x^3-x^2+1` is horizontal, you need to evaluate y coordinates, such that:

`y = 0^3 - 0^2 + 1 => y = 1`

`y = (2/3)^3-(2/3)^2+1 => y = 8/27 - 4/9 + 1`

`y = (8 - 12 + 27)/27 => y = 23/27`

**Hence, evaluating the requested points, under the given conditions, yields **`(0,1), (2/3,23/27).`