What is the tangent of f(x)=x^3-3x^2-8x+9 ?

The tangent of f(x)=x^3-3x^2-8x+9 is y=(3a_{0}^2-6a_{0}-8)x-2a_{0}^3+3a_{0}^2+9

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tangent of f(x)=x^3-3x^2-8x+9

tangent of

f (x) = x^{3} - 3 x^{2} - 8 x + 9

y = (3 a_{0}^{2} - 6 a_{0} - 8) x - 2 a_{0}^{3} + 3 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} - 3 a_{0}^{2} - 8 a_{0} + 9)

Find the slope of

f (x) = x^{3} - 3 x^{2} - 8 x + 9 : \frac{df ( x )}{d x} = 3 x^{2} - 6 x - 8

Compute the slope of

f (x) = x^{3} - 3 x^{2} - 8 x + 9

x = a_{0} : m = 3 a_{0}^{2} - 6 a_{0} - 8

Find the line with slope m=

3 a_{0}^{2} - 6 a_{0} - 8

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0}^{2} - 8 a_{0} + 9) : y = (3 a_{0}^{2} - 6 a_{0} - 8) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 9

y = (3 a_{0}^{2} - 6 a_{0} - 8) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 9

x \to \infty lim (5 + \frac{2}{x})

\int_{2}^{3} (x - 2)^{3} d x

sec^{2} (x) d y + csc (y) d x = 0, y (\frac{π}{4}) = 0

x \to \infty + lim (3 + \frac{2}{x})

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = x^{2} - 2

What is the tangent of f(x)=x^3-3x^2-8x+9 ?
The tangent of f(x)=x^3-3x^2-8x+9 is y=(3a_{0}^2-6a_{0}-8)x-2a_{0}^3+3a_{0}^2+9