lol heres the REAL anwser.

y=x^2-4x+3 ( you have to complete the square here)

y=(x^2-4x+(2)^2-(2)^2)+3

y=(x^2-4x+4-4)+3 Rearrange the brackets to take -4 out

y=(x^2-4x+4)-4+3

y=(x^2-4x+4)-1 Simplify the things in the bracket

y=(x-2)^2-1

Now switch places with x and y

x=(y-2)^2-1 solve for y

x+1=(y-2)^2 square root both sides (square root=Sqrt)

plus/minus Sqrt(x+1)=y-2

y=2 plus/minus Sqrt(x+1) and nope it's not a function

f(x) = x^2-4x-3. Or

y = x^2-4x-3.

For a given value y ,

y=x^2-4x-3. Or

x^2-4x-3-k = 0. Solving for x ,

x = {4 +or- sqrt(4^2+4(3+y)}/2.

x = 2+2sqrt(4+y) Or where when x>2, y = 2+2sqrt(4+x) is the inverse.

x =2-2sqrt(4+y). Or when x <2, y = 2-sqrt(4+x) is the inverse.

One value of y shares double value of x. So the inverse is not a function.