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

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

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

tangent of

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

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x ^{2}}{x - 3} : \frac{df ( x )}{d x} = \frac{x ^{2} - 6 x}{( x - 3 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2}}{x - 3}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

\frac{\partial}{\partial x} (cos (x + 9 y))

y d y = x d x

\frac{d}{d y} (9 - y^{2})

\frac{d}{d x} (2^{x} + 2 cos (\frac{2 x}{2}))

\int 7 x^{3} x^{4} + 6 d x

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