What is the tangent of y=x^4-5x^3+3x^2-4x,(-1,13) ?

The tangent of y=x^4-5x^3+3x^2-4x,(-1,13) is y=-29x-16

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tangent of y=x^4-5x^3+3x^2-4x,(-1,13)

tangent of

y = x^{4} - 5 x^{3} + 3 x^{2} - 4 x, (- 1, 13)

y = - 29 x - 16

Solution steps

Find the slope of

y = x^{4} - 5 x^{3} + 3 x^{2} - 4 x : \frac{d y}{d x} = 4 x^{3} - 15 x^{2} + 6 x - 4

Compute the slope of

y = x^{4} - 5 x^{3} + 3 x^{2} - 4 x

(- 1, 13) : m = - 29

Find the line with slope m=

- 29

and passing through

(- 1, 13) : y = - 29 x - 16

y = - 29 x - 16

\frac{d y}{d x} = \frac{x y + x}{y}

x \to 1000 lim (x^{\frac{2}{3}})

\frac{d}{d x} (\frac{1}{3} x^{3} - 5 x - \frac{4}{x})

\int x \cdot ln^{5} (x) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} + x e^{2 x}

What is the tangent of y=x^4-5x^3+3x^2-4x,(-1,13) ?
The tangent of y=x^4-5x^3+3x^2-4x,(-1,13) is y=-29x-16