What is the tangent of y=2x^3+3x^2-12x+7 ?

The tangent of y=2x^3+3x^2-12x+7 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+7

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tangent of y=2x^3+3x^2-12x+7

tangent of

y = 2 x^{3} + 3 x^{2} - 12 x + 7

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 7

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} + 7)

Find the slope of

y = 2 x^{3} + 3 x^{2} - 12 x + 7 : \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} + 7) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 7

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 7

\frac{d}{d x} (e^{2 x} (2 x + 1))

h \to 0 lim (\frac{49 + h - 7}{h})

2 \cdot sin (3 t)

y = e^{3 x}, y = e^{- 3 x}, x = - 1

\frac{d}{d x} (\frac{e ^{\frac{7 x}{4}}}{1 - 9 x ^{2}})

What is the tangent of y=2x^3+3x^2-12x+7 ?
The tangent of y=2x^3+3x^2-12x+7 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+7