What is the tangent of x^3+3x^2-9x ?

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

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tangent of x^3+3x^2-9x

tangent of

x^{3} + 3 x^{2} - 9 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 3 a_{0}^{2} - 9 a_{0})

Find the slope of

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

Compute the slope of

y = x^{3} + 3 x^{2} - 9 x

x = a_{0} : m = 3 a_{0}^{2} + 6 a_{0} - 9

Find the line with slope m=

3 a_{0}^{2} + 6 a_{0} - 9

and passing through

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

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

\int_{0}^{1} 1 - x^{2} d x

e^{-, at}

\frac{d}{d θ} (sin (\frac{θ}{2}))

\frac{d y}{d x} = \frac{4 x}{y}, y (2) = - 2

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