What is the tangent of 4x^3-13x^2+4 ?

The tangent of 4x^3-13x^2+4 is y=(12a_{0}^2-26a_{0})x-8a_{0}^3+13a_{0}^2+4

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tangent of 4x^3-13x^2+4

tangent

y = (12 a_{0}^{2} - 26 a_{0}) x - 8 a_{0}^{3} + 13 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 13 a_{0}^{2} + 4)

Find the slope of

y = 4 x^{3} - 13 x^{2} + 4 : \frac{d y}{d x} = 12 x^{2} - 26 x

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 = 12 a_{0}^{2} - 26 a_{0}

Find the line with slope m=

12 a_{0}^{2} - 26 a_{0}

and passing through

(a_{0}, 4 a_{0}^{3} - 13 a_{0}^{2} + 4) : y = (12 a_{0}^{2} - 26 a_{0}) x - 8 a_{0}^{3} + 13 a_{0}^{2} + 4

y = (12 a_{0}^{2} - 26 a_{0}) x - 8 a_{0}^{3} + 13 a_{0}^{2} + 4

What is the tangent of 4x^3-13x^2+4 ?
The tangent of 4x^3-13x^2+4 is y=(12a_{0}^2-26a_{0})x-8a_{0}^3+13a_{0}^2+4