What is the tangent of y=2x^3+3x^2-12x+5 ?

The tangent of y=2x^3+3x^2-12x+5 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+5

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tangent of y=2x^3+3x^2-12x+5

tangent of

y = 2 x^{3} + 3 x^{2} - 12 x + 5

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 5

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} + 5)

Find the slope of

y = 2 x^{3} + 3 x^{2} - 12 x + 5 : \frac{d y}{d x} = 6 x^{2} + 6 x - 12

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} - 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} + 5) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 5

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 5

n = 1 \sum \infty e^{\frac{8}{n}} \cdot \frac{1}{n}

3 \frac{d ^{2} y}{d x ^{2}} - 21 \frac{d y}{d x} + 36 y = 5 sin (3 x)

\frac{d}{d x} (\frac{x ^{2} - 81}{x - 9})

\frac{1}{x ^{3}}

\frac{\partial}{\partial y} (2 x y^{2} + 3 x^{2})

What is the tangent of y=2x^3+3x^2-12x+5 ?
The tangent of y=2x^3+3x^2-12x+5 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+5