What is the tangent of y=2x^3-8x ?

The tangent of y=2x^3-8x is y=(6a_{0}^2-8)x-4a_{0}^3

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tangent of y=2x^3-8x

tangent of

y = 2 x^{3} - 8 x

y = (6 a_{0}^{2} - 8) x - 4 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - 8 a_{0})

Find the slope of

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

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}^{2} - 8

Find the line with slope m=

6 a_{0}^{2} - 8

and passing through

(a_{0}, 2 a_{0}^{3} - 8 a_{0}) : y = (6 a_{0}^{2} - 8) x - 4 a_{0}^{3}

y = (6 a_{0}^{2} - 8) x - 4 a_{0}^{3}

What is the tangent of y=2x^3-8x ?
The tangent of y=2x^3-8x is y=(6a_{0}^2-8)x-4a_{0}^3