Question 202334
To find the inverse of the function:
{{{y = x^2-2x+1}}} Exchange the x- and y-variables, then solve the resulting equation for y, thus:
{{{x = y^2-2y+1}}} Solve this equation for y. Subtract X from both sides.
{{{y^2-2y+(1-x) = 0}}} Use the quadratic formula to solve.
{{{y = (-b+-sqrt(b^2-4ac))/2a}}} In the equation just developed, a = 1, b = -2, and c = (1-x).
{{{y = (-(-2+-sqrt(-2)^2-4(1)(1-x)))/2(1)}}} Simplify.
{{{y = (2+-sqrt(4-4+4x))/2}}}
{{{y = (2/2)+sqrt(4x)/2}}} or {{{y = (2/2)-sqrt(4x)/2}}} which can be further simplified into:
{{{highlight(y^-1 = 1+-sqrt(x))}}} This is the inverse.
In order to graph this, you must graph the two solutions as separate graphs:
{{{y = 1+sqrt(x)}}} and {{{y = 1-sqrt(x)}}} 
The graph looks like:
{{{graph(400,400,-5,10,-5,10,x^2-2x+1,1+sqrt(x),1-sqrt(x))}}}
The red parabola is the graph of the given quadratic equation while the blue & green graphs combine to form the graph of the inverse funtion.