What is the normal of y=2x-x^2,(2,0) ?

The normal of y=2x-x^2,(2,0) is y= 1/2 x-1

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normal of y=2x-x^2,(2,0)

normal of

y = 2 x - x^{2}, (2, 0)

y = \frac{1}{2} x - 1

Solution steps

Find the slope of

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

Compute the slope of

y = 2 x - x^{2}

(2, 0) : m = - 2

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(2, 0) : y = \frac{1}{2} x - 1

y = \frac{1}{2} x - 1

\int_{- 1}^{1} x^{- 3} d x

\frac{\partial}{\partial x} (\frac{e ^{x} - e ^{y}}{e ^{x y}})

\frac{\partial}{\partial u} (2 cos (u))

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

\frac{x}{2 x - 1}

What is the normal of y=2x-x^2,(2,0) ?
The normal of y=2x-x^2,(2,0) is y= 1/2 x-1