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

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

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tangent of y=x^2-5x+5

tangent of

y = 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

\frac{d}{d x} (x^{2} + \frac{c}{x ^{2}})

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4)

\int \frac{x ^{2}}{361 - x ^{2}} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 9} e^{3 t}

\int \frac{4 x + 2}{x ^{2} + x + 5} d x

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