Find the point on the graph of y=2x-4 that is closest to the point (1,3). (Optimization equation)

Answer 1

D^2=f(x)= (x-1)^2+((2x-4)-3)^2
f'(x)= 2(x-1)(1)+2(2x-7)(2)
0= 2x-2+(4x-14)2
0= 2x-2+8x-28
0= 10x-30
30=10x
x=3
y=2(3)-4
y=2
(3,2)

the square over D doesn't matter.