What is the tangent of x^3+8 ?

The tangent of x^3+8 is y=3a_{0}^2x-2a_{0}^3+8

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tangent of x^3+8

tangent of

x^{3} + 8

y = 3 a_{0}^{2} x - 2 a_{0}^{3} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 8)

Find the slope of

y = x^{3} + 8 : \frac{d y}{d 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 = 3 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{2}

and passing through

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

y = 3 a_{0}^{2} x - 2 a_{0}^{3} + 8