What is the tangent of y=2x^3+3x^2-12x+6 ?

The tangent of y=2x^3+3x^2-12x+6 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+6

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tangent of y=2x^3+3x^2-12x+6

tangent of

y = 2 x^{3} + 3 x^{2} - 12 x + 6

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

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} - 12 a_{0} + 6)

Find the slope of

y = 2 x^{3} + 3 x^{2} - 12 x + 6 : \frac{d y}{d x} = 6 x^{2} + 6 x - 12

Compute the slope of

y = 2 x^{3} + 3 x^{2} - 12 x + 6

x = a_{0} : m = 6 a_{0}^{2} + 6 a_{0} - 12

Find the line with slope m=

6 a_{0}^{2} + 6 a_{0} - 12

and passing through

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 6) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = 1

x \to 2 lim (- 10 x + (10))

f (x) = 28 x - 0.01 x^{2}

3 x^{2} - 5 y^{2} = 4

2 x^{3} - 12 x^{2} - 30 x + 9

What is the tangent of y=2x^3+3x^2-12x+6 ?
The tangent of y=2x^3+3x^2-12x+6 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+6