curve y = (x^2) - 4x + 5 is tangent perpendiculat to the (line = 2y + x - 7 = 0)

slope of line is -1/2

slope of normal is thus 2

y' = 2x-4 = 2
x = 3
so, our line with slope 2 at (3,2) is

y-2 = 2(x-3)

plot y=x^2-4x+5 and y = 2(x-3)+2 for 0<x<4