What is the tangent of y=3x^2-x^3 ?

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

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tangent of y=3x^2-x^3

tangent of

y = 3 x^{2} - x^{3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - a_{0}^{3})

Find the slope of

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

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 = 6 a_{0} - 3 a_{0}^{2}

Find the line with slope m=

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

and passing through

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

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

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