What is the tangent of x^3-4x^2-9x+9 ?

The tangent of x^3-4x^2-9x+9 is y=(3a_{0}^2-8a_{0}-9)x-2a_{0}^3+4a_{0}^2+9

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tangent of x^3-4x^2-9x+9

tangent of

x^{3} - 4 x^{2} - 9 x + 9

y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 9 a_{0} + 9)

Find the slope of

y = x^{3} - 4 x^{2} - 9 x + 9 : \frac{d y}{d x} = 3 x^{2} - 8 x - 9

Compute the slope of

y = x^{3} - 4 x^{2} - 9 x + 9

x = a_{0} : m = 3 a_{0}^{2} - 8 a_{0} - 9

Find the line with slope m=

3 a_{0}^{2} - 8 a_{0} - 9

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 9 a_{0} + 9) : y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 9

y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 9

\int \frac{1}{4 + ( x - 1 ) ^{2}} d x

\frac{d}{d x} (x^{3} \cdot sin (x))

\int 9 x^{- 1} d x

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

f (x) = \frac{3}{( x + 5 ) ^{2}}, at x = 1

What is the tangent of x^3-4x^2-9x+9 ?
The tangent of x^3-4x^2-9x+9 is y=(3a_{0}^2-8a_{0}-9)x-2a_{0}^3+4a_{0}^2+9