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

\frac{d v}{d t} = - v

\int s h \frac{x}{3} d x

f (x) = 4 x^{2} - 8, at 3

n = 1 \sum \infty \frac{3}{n}

\int \frac{2}{t - t ^{3}} d t

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