What is the tangent of 8x^3-18x^2-6 ?

The tangent of 8x^3-18x^2-6 is y=(24a_{0}^2-36a_{0})x-16a_{0}^3+18a_{0}^2-6

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tangent of 8x^3-18x^2-6

tangent of

8 x^{3} - 18 x^{2} - 6

y = (24 a_{0}^{2} - 36 a_{0}) x - 16 a_{0}^{3} + 18 a_{0}^{2} - 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{3} - 18 a_{0}^{2} - 6)

Find the slope of

y = 8 x^{3} - 18 x^{2} - 6 : \frac{d y}{d x} = 24 x^{2} - 36 x

Compute the slope of

y = 8 x^{3} - 18 x^{2} - 6

x = a_{0} : m = 24 a_{0}^{2} - 36 a_{0}

Find the line with slope m=

24 a_{0}^{2} - 36 a_{0}

and passing through

(a_{0}, 8 a_{0}^{3} - 18 a_{0}^{2} - 6) : y = (24 a_{0}^{2} - 36 a_{0}) x - 16 a_{0}^{3} + 18 a_{0}^{2} - 6

y = (24 a_{0}^{2} - 36 a_{0}) x - 16 a_{0}^{3} + 18 a_{0}^{2} - 6

\int - 2 x + 18 x^{2} - 4 x^{3} + 16 d x

x \to 3 lim (4 x^{2} - 2 x - 6)

y = \frac{x}{x + 9}, (1, 0.1)

x \to \infty lim (- 5 x^{2})

f (x) = \frac{5}{x ^{2}} + sin (x)

What is the tangent of 8x^3-18x^2-6 ?
The tangent of 8x^3-18x^2-6 is y=(24a_{0}^2-36a_{0})x-16a_{0}^3+18a_{0}^2-6