How do you find the inverse of #y = 2x^2 - 3x +1# and is it a function?

1 Answer
Burglar
Jun 5, 2016

To find the inverse you need to write #x=# something, then it is exactly the same way as to solve the second order equation:

#y=2x^2-3x+1#

#2x^2-3x+1-y=0#

#x=(3\pm sqrt(9-4*2*(1-y)))/4#

#x=(3\pm sqrt(1-8y))/4#.

This is the inverse. It is not a function because for one #y# you have two #x#, one with the plus solution and one with the minus.

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