What is the tangent of-3x^3-2x^2-x+3 ?

The tangent of-3x^3-2x^2-x+3 is y=(-9a_{0}^2-4a_{0}-1)x+6a_{0}^3+2a_{0}^2+3

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tangent of-3x^3-2x^2-x+3

tangent of

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

y = (- 9 a_{0}^{2} - 4 a_{0} - 1) x + 6 a_{0}^{3} + 2 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - a_{0} + 3)

Find the slope of

y = - 3 x^{3} - 2 x^{2} - x + 3 : \frac{d y}{d x} = - 9 x^{2} - 4 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 = - 9 a_{0}^{2} - 4 a_{0} - 1

Find the line with slope m=

- 9 a_{0}^{2} - 4 a_{0} - 1

and passing through

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - a_{0} + 3) : y = (- 9 a_{0}^{2} - 4 a_{0} - 1) x + 6 a_{0}^{3} + 2 a_{0}^{2} + 3

y = (- 9 a_{0}^{2} - 4 a_{0} - 1) x + 6 a_{0}^{3} + 2 a_{0}^{2} + 3

x^{2}, 2 x - 1, 0

\frac{\partial}{\partial x} (ln (x^{3} + y^{3}) - z)

\int x x^{2} - 9 d x

\frac{1}{( 1 + x ) ^{\frac{1}{2}}}

\frac{\partial}{\partial t} (\frac{t}{s})

What is the tangent of-3x^3-2x^2-x+3 ?
The tangent of-3x^3-2x^2-x+3 is y=(-9a_{0}^2-4a_{0}-1)x+6a_{0}^3+2a_{0}^2+3