What is the tangent of y=x^2-2x ?

The tangent of y=x^2-2x is y=(2a_{0}-2)x-a_{0}^2

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tangent of y=x^2-2x

tangent of

y = x^{2} - 2 x

y = (2 a_{0} - 2) x - a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 2 a_{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 = a_{0} : m = 2 a_{0} - 2

Find the line with slope m=

2 a_{0} - 2

and passing through

(a_{0}, a_{0}^{2} - 2 a_{0}) : y = (2 a_{0} - 2) x - a_{0}^{2}

y = (2 a_{0} - 2) x - a_{0}^{2}

\int x^{3} 1 + x^{4} d x

f (x) = \frac{1}{x - 1}, at x = 4

\int 4 x sin (2 x^{2}) d x

\frac{\partial}{\partial x} (x^{6} + 8 x y^{5})

\frac{\partial}{\partial x} (3 x + 4 y)

What is the tangent of y=x^2-2x ?
The tangent of y=x^2-2x is y=(2a_{0}-2)x-a_{0}^2