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

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

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

tangent of

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

y = - 4 x

Solution steps

Find the slope of

y = - 3 x^{3} - 3 x^{2} - x + 3 : \frac{d y}{d x} = - 9 x^{2} - 6 x - 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 4) : y = - 4 x

y = - 4 x

y^{^{''}} + 2 y^{^{'}} + 2 = cos (t), y (0) = 0, y^{^{'}} (0) = 0

y = - 3 x \frac{d y}{d x}

(v^{- 3} + 3 v^{- 2})^{3}

\frac{d}{d x} ((9 x + 5) x^{2})

y = 8 - 6 (x - 1)

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