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

Compute the slope of

y = 2 x^{3} - 8 x

(- 2, 0) : m = 16

Find the line with slope m=

16

and passing through

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

y = 16 x + 32

\int e^{- t (s + 1)} d t

\frac{e ^{- s}}{1 - s}

f (t) = 2^{t^{3}}

sin (x), (\frac{π}{4})

y^{^{'''}} - 6 y^{^{''}} = 4 - cos (x)

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