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

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

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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 = - 2 x

Solution steps

Find the tangent point:

(1, - 2)

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 = - 2

Find the line with slope m=

- 2

and passing through

(1, - 2) : y = - 2 x

y = - 2 x

\int 14 ln (3 x) d x

y = \frac{7}{x} + 6 sec (x)

\frac{\partial}{\partial y} (\frac{3 x}{( x + y ) ^{4}})

\int 3 x^{2} - 3 d x

f^{^{'}} (x) = 3 e^{x} + 7 sec^{2} (x)

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