What is the tangent of x^3-6x^2+5x ?

The tangent of x^3-6x^2+5x is y=(3a_{0}^2-12a_{0}+5)x-2a_{0}^3+6a_{0}^2

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tangent of x^3-6x^2+5x

tangent of

x^{3} - 6 x^{2} + 5 x

y = (3 a_{0}^{2} - 12 a_{0} + 5) x - 2 a_{0}^{3} + 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 5 a_{0})

Find the slope of

y = x^{3} - 6 x^{2} + 5 x : \frac{d y}{d x} = 3 x^{2} - 12 x + 5

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} - 12 a_{0} + 5

Find the line with slope m=

3 a_{0}^{2} - 12 a_{0} + 5

and passing through

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 5 a_{0}) : y = (3 a_{0}^{2} - 12 a_{0} + 5) x - 2 a_{0}^{3} + 6 a_{0}^{2}

y = (3 a_{0}^{2} - 12 a_{0} + 5) x - 2 a_{0}^{3} + 6 a_{0}^{2}

\int \frac{1}{2 x - 4} d x

\frac{\partial}{\partial x} (x t)

\int 2 x^{3} + 1 d x

\int csc^{2} (3 x) d x

\frac{1}{7} sin (7 x)

What is the tangent of x^3-6x^2+5x ?
The tangent of x^3-6x^2+5x is y=(3a_{0}^2-12a_{0}+5)x-2a_{0}^3+6a_{0}^2