What is the tangent of f(x)=2x^3-9x^2-60x-2 ?

The tangent of f(x)=2x^3-9x^2-60x-2 is y=(6a_{0}^2-18a_{0}-60)x-4a_{0}^3+9a_{0}^2-2

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tangent of f(x)=2x^3-9x^2-60x-2

tangent of

f (x) = 2 x^{3} - 9 x^{2} - 60 x - 2

y = (6 a_{0}^{2} - 18 a_{0} - 60) x - 4 a_{0}^{3} + 9 a_{0}^{2} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - 9 a_{0}^{2} - 60 a_{0} - 2)

Find the slope of

f (x) = 2 x^{3} - 9 x^{2} - 60 x - 2 : \frac{df ( x )}{d x} = 6 x^{2} - 18 x - 60

Compute the slope of

f (x) = 2 x^{3} - 9 x^{2} - 60 x - 2

x = a_{0} : m = 6 a_{0}^{2} - 18 a_{0} - 60

Find the line with slope m=

6 a_{0}^{2} - 18 a_{0} - 60

and passing through

(a_{0}, 2 a_{0}^{3} - 9 a_{0}^{2} - 60 a_{0} - 2) : y = (6 a_{0}^{2} - 18 a_{0} - 60) x - 4 a_{0}^{3} + 9 a_{0}^{2} - 2

y = (6 a_{0}^{2} - 18 a_{0} - 60) x - 4 a_{0}^{3} + 9 a_{0}^{2} - 2

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

\int csc^{2} (8 θ) cot (8 θ) d θ

f (θ) = cos (θ^{2})

\int_{0}^{1} x^{2} sin (nπ x) d x

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

What is the tangent of f(x)=2x^3-9x^2-60x-2 ?
The tangent of f(x)=2x^3-9x^2-60x-2 is y=(6a_{0}^2-18a_{0}-60)x-4a_{0}^3+9a_{0}^2-2