What is the tangent of-x^3+9x^2-6x+9 ?

The tangent of-x^3+9x^2-6x+9 is y=(-3a_{0}^2+18a_{0}-6)x+2a_{0}^3-9a_{0}^2+9

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tangent of-x^3+9x^2-6x+9

tangent of

- x^{3} + 9 x^{2} - 6 x + 9

y = (- 3 a_{0}^{2} + 18 a_{0} - 6) x + 2 a_{0}^{3} - 9 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - a_{0}^{3} + 9 a_{0}^{2} - 6 a_{0} + 9)

Find the slope of

y = - x^{3} + 9 x^{2} - 6 x + 9 : \frac{d y}{d x} = - 3 x^{2} + 18 x - 6

Compute the slope of

y = - x^{3} + 9 x^{2} - 6 x + 9

x = a_{0} : m = - 3 a_{0}^{2} + 18 a_{0} - 6

Find the line with slope m=

- 3 a_{0}^{2} + 18 a_{0} - 6

and passing through

(a_{0}, - a_{0}^{3} + 9 a_{0}^{2} - 6 a_{0} + 9) : y = (- 3 a_{0}^{2} + 18 a_{0} - 6) x + 2 a_{0}^{3} - 9 a_{0}^{2} + 9

y = (- 3 a_{0}^{2} + 18 a_{0} - 6) x + 2 a_{0}^{3} - 9 a_{0}^{2} + 9

sin (2 x) cos (3 x)

\int x^{2} e^{x} cos (x) d x

\int 2 x cos (x^{2}) d x

f (x) = 0.2 x^{2} + 10, g (x) = x, x = - 7, x = 2

\frac{\partial}{\partial x} (ln (8 x^{2} + 9 y^{2}))

What is the tangent of-x^3+9x^2-6x+9 ?
The tangent of-x^3+9x^2-6x+9 is y=(-3a_{0}^2+18a_{0}-6)x+2a_{0}^3-9a_{0}^2+9