What is the perpendicular \at (sqrt(2)2),y=0 ?

The perpendicular \at (sqrt(2)2),y=0 is x=b

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perpendicular \at (sqrt(2)2),y=0

perpendicular

at (22), y = 0

x = b

Solution steps

Find the slope of

22 : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Line with slope m=

\infty : x = b

x = b

s l o p e in t erce pt - 2 y = 6 + 4 x

mi d p o in t (9, - 10), (- 3, 4)

in f l ec t i o n f (x) = x^{4} - 6 x^{2}

d i s t an ce (- 6, 2), (- 1, 3)

in v erse f (x) = \frac{x + 3}{3}

What is the perpendicular \at (sqrt(2)2),y=0 ?
The perpendicular \at (sqrt(2)2),y=0 is x=b