# determine the equation of the line that passes through (2,2) and is parallel to the line tangent to `y=-3x^3-2x at (-1,5)

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You need to find the equation of the tangent line to the curve `f(x) = y = -3x^3 - 2x ` at the point (-1,5) such that:

`y - f(-1) = f'(-1)(x + 1)`

You need to find f'(x) hence you need to differentiate the function `y = -3x^3 - 2x ` with respect to x such that:

`f'(x) = -9x^2 - 2`

You need to find f'(-1), hence you need to substitute -1 for x in f'(x) such that:

`f'(-1) = -9(-1)^2 - 2 =-9 - 2 = -11`

`y - 5 = -11(x + 1)`

`y = -11x - 11 + 5`

`y = -11x - 6`

You need to remember that the slope of the parallel line to `y = -11x - 6` has the same slope, hence you may write this line using the given point and the slope m = -11 such that:

`y - 2 = -11(x - 2) =gt y = -11x + 22 + 2`

`y = -11x + 24`

**Hence, evaluating the equation of the line that passes through (2,2) and it is parallel to the tangent line to the curve `y = -3x^3 - 2x` at the point (-1,5) yields `y = -11x + 24.` **

How did you get f(x)=-9x^2-2 ?