rearranging f(y) equation to f(x) equation?

Question

Normally I can do this, but with the following I'm lost:

x= sqrt(2y-y^2)

I want y= .....

So far I have y(2-y) = x^2

help?

Answer 1

You have to do what's called completing the square. You were right to square both sides and get x^2=(2y-y^2).

Now multiply both sides by negative 1 so that the coefficient of y^2 is positive. Then you get y^2 - 2y = -x^2.

Now complete the square on the left side (which means you have to find the number to add or subtract that will give you two identical factors). To do this you take -b/2 where b is the coefficient of y. Therefore you get -2/2, which is 1. So if you add one to both sides you get y^2 - 2y +1 = -x^2 +1.

Factor the left side and you get (y-1)(y-1) = -x^2 +1
Which is the same as (y-1)^2 = -x^2 + 1

Take the square root of both sides and you get y-1 = sqrt(-x^2+1).

Isolate the y by adding 1 to both sides and you get y=sqrt(-x^2 +1) +1

Note: Wally A, how do you figure? Did you read my explanation?

And thank you mathsman, I love you, lol.

Answer 2

I don't know what the previous person is on about.
A W 's answer is fine although I would express it as
y = 1 + sqrt(1 - x^2)
Of course the domain must be x between -1 and 1 for y to be real valued, but so what. The original function was restricted to y between 0 and 2 for x to be real valued anyway. Also he talks about equations but we are dealing with functions; not the same thing at all.
You may be interested to hear that in Britian (for no particular reason that I know) someone saying something silly is called
'a right wally'.

Answer 3

Unless the solution is a complex number , it may contain a typo.

Dear A W Math teacher , how did you miss that :D.

an imaginary number must be introduced to complete the solution of the equation as it is.

Answer 4

you must solve the equation
-y^2+2y =x^2 ---> -y^2 +2y -x^2=0
y^2-2y+x^2=0

y = (+2+/- (4-4x^2))/2=0

y = 1+/- (1-x^2)^0.5

this is only definite for -1