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

The tangent of y=x^2-x is y=(2a_{0}-1)x-a_{0}^2

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tangent of y=x^2-x

tangent of

y = x^{2} - x

y = (2 a_{0} - 1) 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} - a_{0})

Find the slope of

y = x^{2} - x : \frac{d y}{d x} = 2 x - 1

Compute the slope of

y = x^{2} - x

x = a_{0} : m = 2 a_{0} - 1

Find the line with slope m=

2 a_{0} - 1

and passing through

(a_{0}, a_{0}^{2} - a_{0}) : y = (2 a_{0} - 1) x - a_{0}^{2}

y = (2 a_{0} - 1) x - a_{0}^{2}

x \to 0 lim (csc (0))

5.7 x + 2.2

\frac{d}{d x} (5 x^{2} + 4 x)

\frac{d}{d x} (2^{10 x})

\int x^{3} 256 - x^{2} d x

What is the tangent of y=x^2-x ?
The tangent of y=x^2-x is y=(2a_{0}-1)x-a_{0}^2