Question 886820
Let's say you want {{{y=f^-1(x)}}}.  Solve for y in {{{y^2-8y+5=x}}}.  You may need to complete the square for y to do it.



{{{y^2-8y+16-16+5=x+16}}}, the 16 completes the missing square
{{{(y-4)^2-16+5=x+16}}}
{{{(y-4)^2-11=x+16}}}
{{{(y-4)^2=x+16+11}}}
{{{(y-4)^2=x+27}}}
{{{y-4=0+- sqrt(x+27)}}}
{{{highlight(y=4+- sqrt(x+27))}}}