What is the tangent of-2x^3-4x^2,\at x=1 ?

The tangent of-2x^3-4x^2,\at x=1 is y=-14x+8

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tangent of-2x^3-4x^2,\at x=1

tangent of

- 2 x^{3} - 4 x^{2}, at x = 1

y = - 14 x + 8

Solution steps

Find the tangent point:

(1, - 6)

Find the slope of

y = - 2 x^{3} - 4 x^{2} : \frac{d y}{d x} = - 6 x^{2} - 8 x

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 = - 14

Find the line with slope m=

- 14

and passing through

(1, - 6) : y = - 14 x + 8

y = - 14 x + 8

\int cot^{2} (x) csc^{4} (x) d x

\frac{\partial}{\partial x} (sin (3 x + 2 y))

\frac{s}{( s ^{2} + 16 ) ^{2}}

4 y^{^{''}} + 6 y^{^{'}} + 7 y = 0

\frac{d}{d x} ((4 x - x^{2})^{100})

What is the tangent of-2x^3-4x^2,\at x=1 ?
The tangent of-2x^3-4x^2,\at x=1 is y=-14x+8