What is the tangent of f(x)=x^3-2x^2-6x+6,\at x=0 ?

The tangent of f(x)=x^3-2x^2-6x+6,\at x=0 is y=-6x+6

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tangent of f(x)=x^3-2x^2-6x+6,\at x=0

tangent of

f (x) = x^{3} - 2 x^{2} - 6 x + 6, at x = 0

y = - 6 x + 6

Solution steps

Find the tangent point:

(0, 6)

Find the slope of

f (x) = x^{3} - 2 x^{2} - 6 x + 6 : \frac{df ( x )}{d x} = 3 x^{2} - 4 x - 6

Compute the slope of

f (x) = x^{3} - 2 x^{2} - 6 x + 6

x = 0 : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 6) : y = - 6 x + 6

y = - 6 x + 6

\int (3 x)^{2} d x

\frac{d}{d x} (- x y^{3})

(- 2, 4), (- 1, - 1)

\int_{- a}^{a} (a - ∣ x ∣) d x

a \to 0 lim (x^{a})

What is the tangent of f(x)=x^3-2x^2-6x+6,\at x=0 ?
The tangent of f(x)=x^3-2x^2-6x+6,\at x=0 is y=-6x+6