What is the tangent of f(x)=x^2-x ?

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

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tangent of f(x)=x^2-x

tangent of

f (x) = 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

f (x) = x^{2} - x : \frac{df ( x )}{d x} = 2 x - 1

Compute the slope of

f (x) = 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}

2 x x

y = x + tan (x), (π, π)

x \to 0 lim (ln (sinh (x)))

\int \frac{4}{369} x^{3} d x

x \to 1 lim (3 x^{2} + 4)

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