Find all points where the tangent line to y = x^3 - 4x + 15 has the slope -1?

You need to remember that at the point of tangency, the slope of the tangent line needs to be equal to the derivative of the function, such that:

`f'(x) = m`

Since the problem provides the value of the slope `m` , you need to plug in the value `-1` in the equation above, such that:

`f'(x) = -1`

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

`f'(x) = (x^3 - 4x + 15)' => f'(x) = 3x^2 - 4`

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

`3x^2 - 4 = -1 => 3x^2 = 4 - 1 => 3x^2 = 3 => x^2 = 1 => x_(1,2) = +-1`

You need to evaluate the coordinates `y_1,y_2,` hence, you need to plug `x_(1,2)` in the equation of the function, such that:

`y_1 = 1^3 - 4*1 + 15 => y_1 = 12`

`y_2 = (-1)^3 - 4*(-1) + 15 => y_2 = -1 + 4 + 15 => y_2 = 18`

Hence, evaluating the points where the tangent line to y = x^3 - 4x + 15 has the slope -1, yields `(1,12) and (-1,18).`

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