What is the tangent of (x+1)(4x^2-4x-8) ?

The tangent of (x+1)(4x^2-4x-8) is y=(12a_{0}^2-12)x-8a_{0}^3-8

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tangent of (x+1)(4x^2-4x-8)

tangent of

(x + 1) (4 x^{2} - 4 x - 8)

y = (12 a_{0}^{2} - 12) x - 8 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} + 1) (4 a_{0}^{2} - 4 a_{0} - 8))

Find the slope of

y = (x + 1) (4 x^{2} - 4 x - 8) : \frac{d y}{d x} = 12 x^{2} - 12

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} - 12

Find the line with slope m=

12 a_{0}^{2} - 12

and passing through

(a_{0}, (a_{0} + 1) (4 a_{0}^{2} - 4 a_{0} - 8)) : y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

What is the tangent of (x+1)(4x^2-4x-8) ?
The tangent of (x+1)(4x^2-4x-8) is y=(12a_{0}^2-12)x-8a_{0}^3-8