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

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

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

tangent of

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

y = - 12 t + 16

Solution steps

Find the slope of

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

Compute the slope of

y = - t^{3}

(2, - 8) : m = - 12

Find the line with slope m=

- 12

and passing through

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

y = - 12 t + 16

x^{2} + x - 2, - 1, 3

f (x) = - \frac{1}{x}, at x = 3

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

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

\int e^{2 x} arcsin (e^{x}) d x

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