What is the tangent of x^3-6x^2-34x+40,\at x=4 ?

The tangent of x^3-6x^2-34x+40,\at x=4 is y=-34x+8

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tangent of x^3-6x^2-34x+40,\at x=4

tangent of

x^{3} - 6 x^{2} - 34 x + 40, at x = 4

y = - 34 x + 8

Solution steps

Find the tangent point:

(4, - 128)

Find the slope of

y = x^{3} - 6 x^{2} - 34 x + 40 : \frac{d y}{d x} = 3 x^{2} - 12 x - 34

Compute the slope of

y = x^{3} - 6 x^{2} - 34 x + 40

x = 4 : m = - 34

Find the line with slope m=

- 34

and passing through

(4, - 128) : y = - 34 x + 8

y = - 34 x + 8

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

\frac{\partial}{\partial z} (\frac{x y}{x + y + z})

\frac{d}{d x} (\frac{cos ( x )}{4 x})

x \to 1 - lim (x^{3} - 4)

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

What is the tangent of x^3-6x^2-34x+40,\at x=4 ?
The tangent of x^3-6x^2-34x+40,\at x=4 is y=-34x+8