What is the tangent of 8x^3-36x^2-96 ?

The tangent of 8x^3-36x^2-96 is y=(24a_{0}^2-72a_{0})x-16a_{0}^3+36a_{0}^2-96

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tangent of 8x^3-36x^2-96

tangent of

8 x^{3} - 36 x^{2} - 96

y = (24 a_{0}^{2} - 72 a_{0}) x - 16 a_{0}^{3} + 36 a_{0}^{2} - 96

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{3} - 36 a_{0}^{2} - 96)

Find the slope of

y = 8 x^{3} - 36 x^{2} - 96 : \frac{d y}{d x} = 24 x^{2} - 72 x

Compute the slope of

y = 8 x^{3} - 36 x^{2} - 96

x = a_{0} : m = 24 a_{0}^{2} - 72 a_{0}

Find the line with slope m=

24 a_{0}^{2} - 72 a_{0}

and passing through

(a_{0}, 8 a_{0}^{3} - 36 a_{0}^{2} - 96) : y = (24 a_{0}^{2} - 72 a_{0}) x - 16 a_{0}^{3} + 36 a_{0}^{2} - 96

y = (24 a_{0}^{2} - 72 a_{0}) x - 16 a_{0}^{3} + 36 a_{0}^{2} - 96

\frac{d}{d x} (2 arcsec (x))

\int e^{5 x} cos (7 x) d x

(3, 8), (7, - 7)

x \to 3 + lim (1 - x)

\int_{1}^{4} 2 π x (e^{x} - ln (x)) d x

What is the tangent of 8x^3-36x^2-96 ?
The tangent of 8x^3-36x^2-96 is y=(24a_{0}^2-72a_{0})x-16a_{0}^3+36a_{0}^2-96