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

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

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

tangent of

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

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

\int 7 sin^{2} (t) cos^{4} (t) d t

\frac{( 2 t e ^{- 3 t} )}{5} + 3 t \cdot sin (t)

(13 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{\partial}{\partial x} (y x + 2 x + g (y))

y^{^{'}} = - x y

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