What is the tangent of f(x)=-x^3+x^2+4x-2,\at x=0 ?

The tangent of f(x)=-x^3+x^2+4x-2,\at x=0 is y=4x-2

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tangent of f(x)=-x^3+x^2+4x-2,\at x=0

tangent of

f (x) = - x^{3} + x^{2} + 4 x - 2, at x = 0

y = 4 x - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

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

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 = 4

Find the line with slope m=

4

and passing through

(0, - 2) : y = 4 x - 2

y = 4 x - 2

What is the tangent of f(x)=-x^3+x^2+4x-2,\at x=0 ?
The tangent of f(x)=-x^3+x^2+4x-2,\at x=0 is y=4x-2