What is the tangent of y=x^3-4x,(2,0) ?

The tangent of y=x^3-4x,(2,0) is y=8x-16

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tangent of y=x^3-4x,(2,0)

tangent of

y = x^{3} - 4 x, (2, 0)

y = 8 x - 16

Solution steps

Find the slope of

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

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 = 8

Find the line with slope m=

8

and passing through

(2, 0) : y = 8 x - 16

y = 8 x - 16

y^{^{''}} + 25 y = - 30 sin (5 t)

\int_{2}^{4} \frac{1}{( x - 1 ) ( x - 4 )} d x

\frac{d}{d x} (2 (1 + cos (x)))

\frac{d}{d x} (\frac{1 - e ^{7 x}}{1 + e ^{7 x}})

y = x^{\frac{8}{5}}

What is the tangent of y=x^3-4x,(2,0) ?
The tangent of y=x^3-4x,(2,0) is y=8x-16