What is the tangent of y=2x^3-8x,(2,0) ?

The tangent of y=2x^3-8x,(2,0) is y=16x-32

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tangent of y=2x^3-8x,(2,0)

tangent of

y = 2 x^{3} - 8 x, (2, 0)

y = 16 x - 32

Solution steps

Find the slope of

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

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 = 16

Find the line with slope m=

16

and passing through

(2, 0) : y = 16 x - 32

y = 16 x - 32

\int (\frac{3 x ^{3} + 2 x}{x}) d x

y = x + 2, y = x^{2}

\frac{d}{d x} (\frac{25 - x ^{2} - y ^{2}}{y})

x \to \frac{π}{2} lim (tan (\frac{x}{2}))

\frac{d}{d x} (f (x) (1 x^{2} - 1))

What is the tangent of y=2x^3-8x,(2,0) ?
The tangent of y=2x^3-8x,(2,0) is y=16x-32