Question 1200940
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The line  2x-y+2 = 0   can be written in  y  = mx+b  (m=slope, b=y-intercept) form:

   y = 2x + 2

By inspection, the slope is 2


A line perpendicular to a line with slope m has slope  -1/m.  So for the above line, a line perpendicular to it has slope -1/2


For the curve {{{ y = (x-2)^2 }}}  you can find the slope at any point by taking the derivative:   {{{ dy/dx = 2(x-2)*1 }}} or  {{{ dy/dx = 2x-4 }}}
The value of the derivative dy/dx at a given x, is the the slope of the tangent of {{{ (x-2)^2 }}} at x.   

You are looking for x such that dy/dx = 2x - 4 = -1/2
Solving for x:  2x = -1/2 + 8/2 = 7/2  --->  x = (7/2)/2 = 7/4

At x = 7/4,  y is {{{ ((7/4)-2)^2 }}} = {{{ ((7/4)-(8/4))^2 }}} = {{{ (-1/4)^2 }}} =  1/16

Thus, the point  (7/4, 1/16) on  {{{ y=(x-2)^2 }}} has a tangent perpendicular to the line  y = 2x + 2.

This illustrates the situation:
  *[illustration graph1]