What is the normal of y= 4/(2x+8),(0, 1/2) ?

The normal of y= 4/(2x+8),(0, 1/2) is y=8x+1/2

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normal of y= 4/(2x+8),(0, 1/2)

normal of

y = \frac{4}{2 x + 8}, (0, \frac{1}{2})

y = 8 x + \frac{1}{2}

Solution steps

Find the slope of

y = \frac{4}{2 x + 8} : \frac{d y}{d x} = - \frac{2}{( x + 4 ) ^{2}}

Compute the slope of

y = \frac{4}{2 x + 8}

(0, \frac{1}{2}) : m = - \frac{1}{8}

Compute the slope of the perpendicular line:

m_{p} = 8

Find the line with slope m=

8

and passing through

(0, \frac{1}{2}) : y = 8 x + \frac{1}{2}

y = 8 x + \frac{1}{2}

\frac{d}{d x} (2 x^{2} + 4 x - 10)

\frac{d}{d x} (\frac{e ^{- x^{3}}}{x})

\frac{1}{s ^{2} + 2 s + 10}

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

\frac{d}{d x} (x + 1 \cdot e^{x})

What is the normal of y= 4/(2x+8),(0, 1/2) ?
The normal of y= 4/(2x+8),(0, 1/2) is y=8x+1/2