What is the tangent of 1/3 x^3-3x^2-16x+8 ?

The tangent of 1/3 x^3-3x^2-16x+8 is y=(a_{0}^2-6a_{0}-16)x-2/3 a_{0}^3+3a_{0}^2+8

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tangent of 1/3 x^3-3x^2-16x+8

tangent of

\frac{1}{3} x^{3} - 3 x^{2} - 16 x + 8

y = (a_{0}^{2} - 6 a_{0} - 16) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 3 a_{0}^{2} - 16 a_{0} + 8)

Find the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} - 16 x + 8 : \frac{d y}{d x} = x^{2} - 6 x - 16

Compute the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} - 16 x + 8

x = a_{0} : m = a_{0}^{2} - 6 a_{0} - 16

Find the line with slope m=

a_{0}^{2} - 6 a_{0} - 16

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 3 a_{0}^{2} - 16 a_{0} + 8) : y = (a_{0}^{2} - 6 a_{0} - 16) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2} + 8

y = (a_{0}^{2} - 6 a_{0} - 16) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2} + 8

y^{^{'}} cot^{2} (x) + tan (y) = 0

\int_{- 3}^{3} x^{2} d x

y^{^{'}} = - 1 - 4 x + 2 y, y (0) = 5

\frac{d}{d x} (\frac{2}{( 2 - x ) ^{3}})

\int z cos (x z) d x

What is the tangent of 1/3 x^3-3x^2-16x+8 ?
The tangent of 1/3 x^3-3x^2-16x+8 is y=(a_{0}^2-6a_{0}-16)x-2/3 a_{0}^3+3a_{0}^2+8