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

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

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

tangent of

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

y = - 6 t - 4

Solution steps

Find the slope of

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

Compute the slope of

y = - 2 t^{3}

(- 1, 2) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 1, 2) : y = - 6 t - 4

y = - 6 t - 4

\frac{\partial}{\partial y} (\frac{3 x y}{x ^{4} + y ^{4}})

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

x, - 2 x + 3, 0, 1.5

\frac{d}{d x} (5 cos^{2} (x) - 10 sin (x))

\int (x + 1)^{100} d x

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