What is the normal of y= 1/x ,(2, 1/2) ?

The normal of y= 1/x ,(2, 1/2) is y=4x-15/2

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normal of y= 1/x ,(2, 1/2)

normal of

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

y = 4 x - \frac{15}{2}

Solution steps

Find the slope of

y = \frac{1}{x} : \frac{d y}{d x} = - \frac{1}{x ^{2}}

Compute the slope of

y = \frac{1}{x}

(2, \frac{1}{2}) : m = - \frac{1}{4}

Compute the slope of the perpendicular line:

m_{p} = 4

Find the line with slope m=

4

and passing through

(2, \frac{1}{2}) : y = 4 x - \frac{15}{2}

y = 4 x - \frac{15}{2}

\frac{7}{s ^{2} + 49}

\frac{d}{d x} (arcsec (3 x))

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

\int (1 - x)^{2} d x

\int (5 z^{2} - 3)^{3} z d z

What is the normal of y= 1/x ,(2, 1/2) ?
The normal of y= 1/x ,(2, 1/2) is y=4x-15/2