What is the tangent of 2x^3+3x^2-180x-3 ?

The tangent of 2x^3+3x^2-180x-3 is y=(6a_{0}^2+6a_{0}-180)x-4a_{0}^3-3a_{0}^2-3

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tangent of 2x^3+3x^2-180x-3

tangent of

2 x^{3} + 3 x^{2} - 180 x - 3

y = (6 a_{0}^{2} + 6 a_{0} - 180) x - 4 a_{0}^{3} - 3 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 180 a_{0} - 3)

Find the slope of

y = 2 x^{3} + 3 x^{2} - 180 x - 3 : \frac{d y}{d x} = 6 x^{2} + 6 x - 180

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6 a_{0}^{2} + 6 a_{0} - 180

Find the line with slope m=

6 a_{0}^{2} + 6 a_{0} - 180

and passing through

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 180 a_{0} - 3) : y = (6 a_{0}^{2} + 6 a_{0} - 180) x - 4 a_{0}^{3} - 3 a_{0}^{2} - 3

y = (6 a_{0}^{2} + 6 a_{0} - 180) x - 4 a_{0}^{3} - 3 a_{0}^{2} - 3

x \to + 0 + lim (3 x)

f (x) = x^{2} 4 x - 1

g (x) = \frac{1}{4 x ^{2} + 5 x}

sin (5 t - \frac{π}{3}) \cdot 1 (5 t - \frac{π}{3})

\frac{3 z - 5}{z ^{2} - 4 z + 3}

What is the tangent of 2x^3+3x^2-180x-3 ?
The tangent of 2x^3+3x^2-180x-3 is y=(6a_{0}^2+6a_{0}-180)x-4a_{0}^3-3a_{0}^2-3