How to find the inverse of #x^(2)-x-2#?

1 Answer
Feb 11, 2018

#y=1/2+-sqrt(x+9/4)#

Explanation:

start with your equation.
#y=x^2-x-2#

switch #x# and #y#*, then move all y terms to one side and anything else to the other side.
#x=y^2-y-2#
#x+2=y^2-y#

now, solve for y by completing the square.
#x+2=(y^2-y+1/4)-1/4# #rarr# you subtract #1/4# to keep the equation equal
#x+2=(y-1/2)^2-1/4# #rarr# simplify#(y^2-y+1/4)#
#x+9/4=(y-1/2)^2# #rarr# add #1/4# to both sides to keep it equal
#+-sqrt(x+9/4)=y-1/2# #rarr# remember to ALWAYS have the positive and negative sign when square rooting any terms
#y=1/2+-sqrt(x+9/4)# #rarr# add #1/2# to both sides to keep it equal

*You switch #x# and #y# because when you find the inverse, you are finding the equation of the graph flipped over the line #y=x#. This line is also on the graph linked below.

**However, since this function is not one-to-one (each y-value only has one possible x-value), there is technically no real inverse function.

All equations on a graph:
Desmos