How do I find the inverse of #y=(x+3)^2#?

1 Answer
Feb 18, 2018

See the explanation below.

Explanation:

The quadratic function which is represented by the graph of a Parabola fails the Horizontal Line Test thus is not a one to one function. Therefore the inverse is not a function unless with restricted domain.

Edit: Adding the solution below to address update in the comments:

#y=(x+3)^2#
#x=(y+3)^2#
#+-sqrtx=y+3#
#y=-3+-sqrtx# => inverse relation

Restricting the domain: Recall that the domain and the range of the inverse function are the range and the domain of the original function respectively. The range of the original function in this case is #y>=0#, that should be the domain of the inverse function so for the inverse function we take the positive square root as follow:
#f^-1(x)=-3+sqrtx# => this is the inverse function the domain and the range are:
domain: #x>=0#
range: #y>=-3#
This means we need to restrict the domain of the original function to: #x>=-3# in order for it to have an inverse function.