Given:
#color(white)("XXX")y=4-x^2+3x#
Re-arrange the equation to solve for #x# in terms of #y#
#color(white)("XXX")-y=x^2-3x-4#
#color(white)("XXX")-y= (x^2-3xcolor(red)(+(3/2)^2))-4color(red)(-9/4)#
#color(white)("XXX")-y=(x-3/2)^2-25/4#
#color(white)("XXX")(x-3/2)^2=25/4-y#
#color(white)("XXX")x-3/2=+-sqrt(25-4y)/2#
#color(white)("XXX")x=(3+-sqrt(25-4y))/2#
In order to maintain #y# as the dependent variable (and #x# as the independent variable) it is common to exchange the #x# and #y# variables at this point.
In this case the inverse would be written as:
#color(white)("XXX")y=(3+-sqrt(25-x))/2#
Sometimes the #y# is written in some modified form such as #y'# to avoid confusing it with the #y# in the original equation.
Check with your instructor for the form she/he wants you to use.
