# The equation of a curve is y = x^3-x^2-2x-3. find the tangent for this curve at point (3,9)

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jeew-m | Certified Educator

The gradient of the tangent line to any curve is given by the first derivative of the curve.

`y = x^3-x^2-2x-3`

`y' = 3x^2-2x-2`

So the gradient of any tangent line to the curve is given by `3x^2-2x-2` .

`m = 3x^2-2x-2`

But this tangent suppose to go over the point (3,9).

`m = (3xx9-2xx3-2) = 19`

The equation of a tangent is straight line which is y = mx+c

`y = 19x+c`

This line passes through (3,9)

`9 = 19xx3+c`

`c = 48`

*So the equation of the tangent is y = 18x-48*

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