What is the tangent of y=-x^3,(1,-1) ?

The tangent of y=-x^3,(1,-1) is y=-3x+2

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tangent of y=-x^3,(1,-1)

tangent of

y = - x^{3}, (1, - 1)

y = - 3 x + 2

Solution steps

Find the slope of

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

Compute the slope of

y = - x^{3}

(1, - 1) : m = - 3

Find the line with slope m=

- 3

and passing through

(1, - 1) : y = - 3 x + 2

y = - 3 x + 2

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