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

The tangent of f(x)=-x^3+2x^2-4,\at x=-1 is y=-7x-8

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tangent of f(x)=-x^3+2x^2-4,\at x=-1

tangent of

f (x) = - x^{3} + 2 x^{2} - 4, at x = - 1

y = - 7 x - 8

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

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

Compute the slope of

f (x) = - x^{3} + 2 x^{2} - 4

x = - 1 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 1, - 1) : y = - 7 x - 8

y = - 7 x - 8

\frac{d}{d x} (x^{2} - 2 x y + y^{2})

(3 x^{7} - 4) (2 x - 5 x^{2})

x \to - 2 lim (1 + x)

x^{5} + 8

What is the tangent of f(x)=-x^3+2x^2-4,\at x=-1 ?
The tangent of f(x)=-x^3+2x^2-4,\at x=-1 is y=-7x-8