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

The tangent of f(x)=-x^3-3x^2-3,\at x=-3 is y=-9x-30

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

tangent of

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

y = - 9 x - 30

Solution steps

Find the tangent point:

(- 3, - 3)

Find the slope of

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

Compute the slope of

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

x = - 3 : m = - 9

Find the line with slope m=

- 9

and passing through

(- 3, - 3) : y = - 9 x - 30

y = - 9 x - 30

x \to - 1 lim (\frac{x + 2}{x ^{2} + x})

\int (2.286 x - 2.5 (x - 3) - 3 (x - 5)) d x

- 6 e^{- x^{2}} x

\int \frac{1}{x} d x

y = x^{2} + 4, y = 7 - 2 x^{2}

What is the tangent of f(x)=-x^3-3x^2-3,\at x=-3 ?
The tangent of f(x)=-x^3-3x^2-3,\at x=-3 is y=-9x-30