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

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

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

tangent of

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

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = x^{3} - 4 x^{2} : \frac{df ( x )}{d x} = 3 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 = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

d er i v a t i v e ln (\frac{3 + e ^{x}}{3 - e ^{x}})

\frac{d y}{d x} - \frac{1}{x} y = 1 - \frac{1}{x}

d er i v a t i v e f (x) = cos (6 x)

d er i v a t i v e \frac{tan ( 3 x )}{x ^{2}}

\int \frac{24 x ^{6} - 24 x ^{4} + x}{x} d x

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