What is the tangent of x^2-4x+1 ?

The tangent of x^2-4x+1 is y=(2a_{0}-4)x-a_{0}^2+1

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tangent of x^2-4x+1

tangent of

x^{2} - 4 x + 1

y = (2 a_{0} - 4) 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} - 4 a_{0} + 1)

Find the slope of

y = x^{2} - 4 x + 1 : \frac{d y}{d x} = 2 x - 4

Compute the slope of

y = x^{2} - 4 x + 1

x = a_{0} : m = 2 a_{0} - 4

Find the line with slope m=

2 a_{0} - 4

and passing through

(a_{0}, a_{0}^{2} - 4 a_{0} + 1) : y = (2 a_{0} - 4) x - a_{0}^{2} + 1

y = (2 a_{0} - 4) x - a_{0}^{2} + 1

\int sin (x) cos^{6} (x) d x

x \to 0 lim (cos^{2} (x))

\frac{x ^{2} + 1}{x - 2}

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

8 x^{\frac{3}{2}}

What is the tangent of x^2-4x+1 ?
The tangent of x^2-4x+1 is y=(2a_{0}-4)x-a_{0}^2+1