What is the tangent of y=(x^3-16x)^{12},(-4,0) ?

The tangent of y=(x^3-16x)^{12},(-4,0) is y=0

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tangent of y=(x^3-16x)^{12},(-4,0)

tangent of

y = (x^{3} - 16 x)^{12}, (- 4, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 16 x)^{12} : \frac{d y}{d x} = 12 (x^{3} - 16 x)^{11} (3 x^{2} - 16)

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 = 0

Find the line with slope m=

0

and passing through

(- 4, 0) : y = 0

y = 0

\frac{5 x ^{2}}{x}

y^{^{'}} + y = e^{- x}

\frac{8 - sec ( x )}{tan ( x )}

\int \frac{e ^{- y}}{y} d y

\int_{0}^{2} (x - 1)^{2} d x

What is the tangent of y=(x^3-16x)^{12},(-4,0) ?
The tangent of y=(x^3-16x)^{12},(-4,0) is y=0