What is the tangent of f(x)=x^3-8x^2,\at x=0 ?

The tangent of f(x)=x^3-8x^2,\at x=0 is y=0

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tangent of f(x)=x^3-8x^2,\at x=0

tangent of

f (x) = x^{3} - 8 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = x^{3} - 8 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 16 x

Compute the slope of

f (x) = x^{3} - 8 x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

\int sec^{6} (x) tan^{5} (x) d x

x \to 0 lim (\frac{x ^{2}}{2})

\frac{d}{d x} (\frac{( x ^{2} + 3 x - 4 )}{x + 2})

\frac{d}{d x} (- cos (x) - sin (x))

\frac{1}{y} y^{^{'}} = - \frac{1}{x}

What is the tangent of f(x)=x^3-8x^2,\at x=0 ?
The tangent of f(x)=x^3-8x^2,\at x=0 is y=0