What is the tangent of y=-2x^3,(1,-2) ?

The tangent of y=-2x^3,(1,-2) is y=-6x+4

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

tangent of

y = - 2 x^{3}, (1, - 2)

y = - 6 x + 4

Solution steps

Find the slope of

y = - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2}

Compute the slope of

y = - 2 x^{3}

(1, - 2) : m = - 6

Find the line with slope m=

- 6

and passing through

(1, - 2) : y = - 6 x + 4

y = - 6 x + 4

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

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))

\frac{d y}{d x}, y = 1 1^{x}

\frac{d}{d x} ((x^{2} - 3) e^{x})

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

What is the tangent of y=-2x^3,(1,-2) ?
The tangent of y=-2x^3,(1,-2) is y=-6x+4