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

The tangent of y=-x^3,(2,-8) is y=-12x+16

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

tangent of

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

y = - 12 x + 16

Solution steps

Find the slope of

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

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 = - 12

Find the line with slope m=

- 12

and passing through

(2, - 8) : y = - 12 x + 16

y = - 12 x + 16

\frac{d y}{d x} = - (1 + y)^{4}

y^{^{''}} = sin (3 x - \frac{π}{3})

7 e^{- 4 x}

\int \frac{( e ^{x} + 9 )}{e ^{2 x} - 6 e ^{x} + 9} d x

f (x) = x^{2} + 3, (- 3, 12)

What is the tangent of y=-x^3,(2,-8) ?
The tangent of y=-x^3,(2,-8) is y=-12x+16