What is the tangent of 4x^3-18x^2-48x ?

The tangent of 4x^3-18x^2-48x is y=(12a_{0}^2-36a_{0}-48)x-8a_{0}^3+18a_{0}^2

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tangent of 4x^3-18x^2-48x

tangent of

4 x^{3} - 18 x^{2} - 48 x

y = (12 a_{0}^{2} - 36 a_{0} - 48) x - 8 a_{0}^{3} + 18 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 18 a_{0}^{2} - 48 a_{0})

Find the slope of

y = 4 x^{3} - 18 x^{2} - 48 x : \frac{d y}{d x} = 12 x^{2} - 36 x - 48

Compute the slope of

y = 4 x^{3} - 18 x^{2} - 48 x

x = a_{0} : m = 12 a_{0}^{2} - 36 a_{0} - 48

Find the line with slope m=

12 a_{0}^{2} - 36 a_{0} - 48

and passing through

(a_{0}, 4 a_{0}^{3} - 18 a_{0}^{2} - 48 a_{0}) : y = (12 a_{0}^{2} - 36 a_{0} - 48) x - 8 a_{0}^{3} + 18 a_{0}^{2}

y = (12 a_{0}^{2} - 36 a_{0} - 48) x - 8 a_{0}^{3} + 18 a_{0}^{2}

\frac{d}{d x} (4 - 2 x)

2 y = x, \frac{y ^{2}}{2} = x

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

x \to \infty lim (\frac{1}{- x})

\int (2 x^{3} - \frac{2}{3} x^{2} + 5 x) d x

What is the tangent of 4x^3-18x^2-48x ?
The tangent of 4x^3-18x^2-48x is y=(12a_{0}^2-36a_{0}-48)x-8a_{0}^3+18a_{0}^2