What is the tangent of f(x)=-x^3,(2,-8) ?

The tangent of f(x)=-x^3,(2,-8) is y=-12x+16

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tangent of f(x)=-x^3,(2,-8)

tangent of

f (x) = - x^{3}, (2, - 8)

y = - 12 x + 16

Solution steps

Find the slope of

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

Compute the slope of

f (x) = - x^{3}

(2, - 8) : m = - 12

Find the line with slope m=

- 12

and passing through

(2, - 8) : y = - 12 x + 16

y = - 12 x + 16

x \to 0 lim (tan (x))

\int_{- \infty}^{- 1} e^{- 30 t} d t

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

\frac{d}{d x} (lo g_{e} (2))

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

What is the tangent of f(x)=-x^3,(2,-8) ?
The tangent of f(x)=-x^3,(2,-8) is y=-12x+16