What is the tangent of x^3-3x^2+2x-2,\at x=0 ?

The tangent of x^3-3x^2+2x-2,\at x=0 is y=2x-2

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tangent of x^3-3x^2+2x-2,\at x=0

tangent of

x^{3} - 3 x^{2} + 2 x - 2, at x = 0

y = 2 x - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

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

Find the line with slope m=

2

and passing through

(0, - 2) : y = 2 x - 2

y = 2 x - 2

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 2 y})

y = \frac{1}{2 x - 9}

\frac{d}{d x} (e^{- i x})

x \to 0 lim ((1 + x) \frac{1}{x})

\int \frac{15}{1 + 9 x ^{2}} d x

What is the tangent of x^3-3x^2+2x-2,\at x=0 ?
The tangent of x^3-3x^2+2x-2,\at x=0 is y=2x-2