What is the tangent of y=x^3-4x,(-2,0) ?

The tangent of y=x^3-4x,(-2,0) is y=8x+16

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tangent of y=x^3-4x,(-2,0)

tangent of

y = x^{3} - 4 x, (- 2, 0)

y = 8 x + 16

Solution steps

Find the slope of

y = x^{3} - 4 x : \frac{d y}{d x} = 3 x^{2} - 4

Compute the slope of

y = x^{3} - 4 x

(- 2, 0) : m = 8

Find the line with slope m=

8

and passing through

(- 2, 0) : y = 8 x + 16

y = 8 x + 16

\int_{1}^{e^{2}} \frac{ln ^{8} ( x ^{2} )}{x} d x

5 e^{x}

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

y = 3 x^{2} + 3, (2, 15)

\frac{\partial}{\partial x} (e^{k x - m t})

What is the tangent of y=x^3-4x,(-2,0) ?
The tangent of y=x^3-4x,(-2,0) is y=8x+16