What is the tangent of f(x)=x^2-2x+1 ?

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

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tangent of f(x)=x^2-2x+1

tangent of

f (x) = x^{2} - 2 x + 1

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

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} + 1)

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 2 x + 1

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

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

\frac{d}{d x} ((2 x^{2} - 4 x)^{- 2})

x \to \infty lim ((x)^{2})

k = 0 \sum \infty \frac{3 ^{k}}{k !}

\frac{x ^{2} - 3 x + 11}{x ^{2} + x - 3}

\frac{\partial}{\partial θ} (cos (θ))

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