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)=(
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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) (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 )
\ =\ \sqrt{\frac{x\ -\ 5}{-4}}\ +\ 3)
From there, I would pretty it up a little, mostly so that the domain of the inverse function is more obvious:
\ =\ \frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3)
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:
\LARGE}^{-1}(x)\ =\ -\frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3)
But we could NOT have done
\LARGE}^{-1}(x)\ =\ \pm\frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3)
Why? Because \LARGE}^{-1}(x))
is NOT a function. Why? That question is an exercise for the student.
John
My calculator said it, I believe it, that settles it
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