What is the tangent of f(x)=-x^3+2x^2-2,\at x=-1 ?

The tangent of f(x)=-x^3+2x^2-2,\at x=-1 is y=-7x-6

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tangent of f(x)=-x^3+2x^2-2,\at x=-1

tangent of

f (x) = - x^{3} + 2 x^{2} - 2, at x = - 1

y = - 7 x - 6

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = - x^{3} + 2 x^{2} - 2 : \frac{df ( x )}{d x} = - 3 x^{2} + 4 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 = - 7

Find the line with slope m=

- 7

and passing through

(- 1, 1) : y = - 7 x - 6

y = - 7 x - 6

\frac{d y}{d x} + 2 y - 1 = 0

\frac{\partial}{\partial x} (x^{4} y - 4 x^{3} y^{2})

\int (\frac{3 x}{2} - \frac{9}{4}) e^{x (x - 3)} d x

f (x) = 1 - 2 tan (x), at x = \frac{3 π}{4}

y = x^{3} - 15 x

What is the tangent of f(x)=-x^3+2x^2-2,\at x=-1 ?
The tangent of f(x)=-x^3+2x^2-2,\at x=-1 is y=-7x-6