What is the normal of y=x^2-4x,(5,5) ?

The normal of y=x^2-4x,(5,5) is y=-1/6 x+35/6

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normal of y=x^2-4x,(5,5)

normal of

y = x^{2} - 4 x, (5, 5)

y = - \frac{1}{6} x + \frac{35}{6}

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 4 x

(5, 5) : m = 6

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(5, 5) : y = - \frac{1}{6} x + \frac{35}{6}

y = - \frac{1}{6} x + \frac{35}{6}

\frac{\partial}{\partial x} (x^{2} e^{yz})

\int \frac{sin ( lo g _{10} ( x ) )}{x} d x

\int \frac{1}{8} x^{2} d x

\int_{0}^{\infty} t e^{- s t} d t

What is the normal of y=x^2-4x,(5,5) ?
The normal of y=x^2-4x,(5,5) is y=-1/6 x+35/6