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

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

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tangent of y=2x^3-x^2+2

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

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} - 2 a_{0}

Find the line with slope m=

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

and passing through

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

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

\int \frac{e ^{x^{n}}}{x ^{1 - n}} d x

h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{3 h})

\frac{e ^{- s}}{( s ^{2} + 8 )}

\frac{d}{d x} (\frac{3 x - 1}{2 x + 1})

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

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