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   you can find the slope at any point by taking the derivative:    or  

The value of the derivative dy/dx at a given x, is the the slope of the tangent of  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  =  =  =  1/16



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



This illustrates the situation:

  

  

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