What is the tangent of x^2-5x+5 ?

The tangent of x^2-5x+5 is y=(2a_{0}-5)x-a_{0}^2+5

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tangent of x^2-5x+5

tangent of

x^{2} - 5 x + 5

y = (2 a_{0} - 5) x - a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 5 a_{0} + 5)

Find the slope of

y = x^{2} - 5 x + 5 : \frac{d y}{d x} = 2 x - 5

Compute the slope of

y = x^{2} - 5 x + 5

x = a_{0} : m = 2 a_{0} - 5

Find the line with slope m=

2 a_{0} - 5

and passing through

(a_{0}, a_{0}^{2} - 5 a_{0} + 5) : y = (2 a_{0} - 5) x - a_{0}^{2} + 5

y = (2 a_{0} - 5) x - a_{0}^{2} + 5

\int x^{3} x^{2} + 8 d x

\frac{d}{d x} (\frac{1}{arctan ( x ^{2} )})

\int cos^{2} (x) d x

(1 + cos (t)) \frac{d y}{d t} = (1 + e^{- y}) sin (t)

\frac{d}{d x} ((\frac{2 x + 1}{3 x - 1})^{4})

What is the tangent of x^2-5x+5 ?
The tangent of x^2-5x+5 is y=(2a_{0}-5)x-a_{0}^2+5