



 
Hi Shona, I would complete the square on the right side of $y= x^2  2x$ to get $y + 1 = x^2  2x + 1.$ Hence $y + 1 = (x  1)^2 .$ Solving for $x$ gives $x = 1 \pm \sqrt{y + 1}.$ Notice that this inverse is not a function. Penny  


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