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

The tangent of y=-2x^3,(-2,16) is y=-24x-32

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tangent of y=-2x^3,(-2,16)

tangent of

y = - 2 x^{3}, (- 2, 16)

y = - 24 x - 32

Solution steps

Find the slope of

y = - 2 x^{3} : \frac{d y}{d x} = - 6 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 = - 24

Find the line with slope m=

- 24

and passing through

(- 2, 16) : y = - 24 x - 32

y = - 24 x - 32

\int \frac{2 y}{3} d y

t an g e n t f (x) = x^{4} + 4 x^{2} - x, at x = 1

\int sin^{5} (7 x) cos^{2} (7 x) d x

\frac{d}{d x} (2 x - 3 3 x)

\frac{d}{d x} (a sin (x) + b cos (x))

What is the tangent of y=-2x^3,(-2,16) ?
The tangent of y=-2x^3,(-2,16) is y=-24x-32