What is the tangent of y=x^2-2x,\at x=0 ?

The tangent of y=x^2-2x,\at x=0 is y=-2x

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tangent of y=x^2-2x,\at x=0

tangent of

y = x^{2} - 2 x, at x = 0

y = - 2 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = x^{2} - 2 x : \frac{d y}{d x} = 2 x - 2

Compute the slope of

y = x^{2} - 2 x

x = 0 : m = - 2

Find the line with slope m=

- 2

and passing through

(0, 0) : y = - 2 x

y = - 2 x

\int \frac{ln ( x )}{x 1 + 7 ln ^{2} ( x )} d x

\frac{d y}{d x} = 2 y - 7 y^{3}

\frac{d}{d x} ((x)^{5 x})

\int \frac{x ^{3} + x + 4}{x ^{2}} d x

\frac{d}{d y} (arctan (\frac{x}{y}))

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