What is the tangent of y=8+4x^2-2x^3 ?

The tangent of y=8+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+8

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tangent of y=8+4x^2-2x^3

tangent of

y = 8 + 4 x^{2} - 2 x^{3}

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 + 4 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

y = 8 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

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} + 8 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 8 a_{0}

and passing through

(a_{0}, 8 + 4 a_{0}^{2} - 2 a_{0}^{3}) : y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 8

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 8

6 (2 e^{x^{2}} x^{2} + e^{x^{2}})

3 + x

x \to + 0 + lim (ln (2 x))

\int_{1}^{2} \frac{x}{2} d x

\int t t^{2} + 2 d t

What is the tangent of y=8+4x^2-2x^3 ?
The tangent of y=8+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+8