|
||||||||||||
|
||||||||||||
| ||||||||||||
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 | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |