What is the tangent of y=2x^3+3x^2-12x+8 ?

The tangent of y=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 y=2x^3+3x^2-12x+8

tangent of

y = 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

y = 2 x^{3} + 3 x^{2} - 12 x + 8 : \frac{d y}{d x} = 6 x^{2} + 6 x - 12

Compute the slope of

y = 2 x^{3} + 3 x^{2} - 12 x + 8

x = a_{0} : 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

\frac{d y}{d x} - 2 y = 3 x

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{8}})

\int \frac{- 2}{x ^{4}} d x

\frac{d}{d x} (a x^{2} e^{x})

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{2})

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