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

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

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

tangent of

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

y = - 12 x - 16

Solution steps

Find the slope of

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

Compute the slope of

y = - 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 (arctan (x))

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

\frac{1}{( 1 - x )}

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