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