What is the tangent of f(x)=2x^3+3x^2-12x+8 ?

The tangent of f(x)=2x^3+3x^2-12x+8 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+8

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tangent of f(x)=2x^3+3x^2-12x+8

tangent of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 8)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8 : \frac{df ( x )}{d x} = 6 x^{2} + 6 x - 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 = 6 a_{0}^{2} + 6 a_{0} - 12

Find the line with slope m=

6 a_{0}^{2} + 6 a_{0} - 12

and passing through

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 8) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 8

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 8

y = \frac{x ^{x}}{sin ( x )}

x \to 2 lim (\frac{x ^{3} + 8}{x - 2})

\frac{\partial}{\partial y} (e^{x - 1} cos (y))

\frac{\partial}{\partial x} (72 y^{7} x^{7} - 56 y^{6} x^{6})

\frac{1 - x}{1 + x}

What is the tangent of f(x)=2x^3+3x^2-12x+8 ?
The tangent of f(x)=2x^3+3x^2-12x+8 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+8