SOLUTION: I am trying to find the inverse of the following function: f(x) =-4(x-3)^2+5 This is what I have so far(I don't think it is right): y=-4(x-3)^2+5 y-5= =-4(x-3)^2 (y-5)/(-4)=(

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am trying to find the inverse of the following function: f(x) =-4(x-3)^2+5 This is what I have so far(I don't think it is right): y=-4(x-3)^2+5 y-5= =-4(x-3)^2 (y-5)/(-4)=(      Log On


   



Question 627712: I am trying to find the inverse of the following function: f(x) =-4(x-3)^2+5
This is what I have so far(I don't think it is right):
y=-4(x-3)^2+5
y-5= =-4(x-3)^2
(y-5)/(-4)=(x-3)^2
take the sqrt both sides and get
sqrt of (y-5)/(-4)= x-3
x= sqrt of (y-5)/(-4)+3
When I graph this it is wrong...What am I doing wrong?
Thanks for the help!

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You forgot to do the last step in the process of finding the inverse function, namely to swap the variables.



Just like you had it, but then swap and



And then replace with




From there, I would pretty it up a little, mostly so that the domain of the inverse function is more obvious:



But it is still the same function and should make a tidy graph that is a reflection of half of your original graph right across the line . Why only half? Because way back when we took the square root of both sides, we only considered the positive half. We could have just as easily done:



But we could NOT have done



Why? Because is NOT a function. Why? That question is an exercise for the student.

John

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