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