What is the tangent of x^2-18x+1 ?

The tangent of x^2-18x+1 is y=(2a_{0}-18)x-a_{0}^2+1

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tangent of x^2-18x+1

tangent of

x^{2} - 18 x + 1

y = (2 a_{0} - 18) 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} - 18 a_{0} + 1)

Find the slope of

y = x^{2} - 18 x + 1 : \frac{d y}{d x} = 2 x - 18

Compute the slope of

y = x^{2} - 18 x + 1

x = a_{0} : m = 2 a_{0} - 18

Find the line with slope m=

2 a_{0} - 18

and passing through

(a_{0}, a_{0}^{2} - 18 a_{0} + 1) : y = (2 a_{0} - 18) x - a_{0}^{2} + 1

y = (2 a_{0} - 18) x - a_{0}^{2} + 1

\frac{6 s ^{2} + 8}{s ( s ^{2} + 4 )}

\frac{d y}{d x}, y^{2} cos (\frac{1}{y}) = 2 x + 2 y

\int \frac{4 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 45} d x

\frac{d}{d x} ((\frac{5}{x} + x) (\frac{5}{x} - x))

y^{^{'}} = (x - 2) e^{- 2 y}

What is the tangent of x^2-18x+1 ?
The tangent of x^2-18x+1 is y=(2a_{0}-18)x-a_{0}^2+1