Find f(x) given the equation of the curve

[awinzler]

The instructions are "Find a function whose graph is the given curve: the bottom half of the parabola

x + (y - 1)^2 = 0

here is my work:

(y-1)^2 = -x
y-1 = sqrt(-x)
y = 1 + sqrt(-x) = f(x) where x must be negative or zero for sqrt(-x) to be a real number

or

x = -(y-1)^2
sqrt(x) = -y + 1
sqrt(x) + y = 1
y = 1 - sqrt(x) where x must be positive or zero for sqrt(x) to be a real number

the solution manual for the book says f(x) = 1 - sqrt(-x), which is different than either of my answers. which of these three is correct and why? I have worked this problem 8 or 10 times with the same above results and I cant figure out what's wrong. thank you for any help.

 

[mmm4444bot]

Your description of the graph is inadequate.

Graph the three functions. Then you will know which one (if any) matches.

 

[awinzler]

no graph is given in the problem, only the equation for the curve. i have provided ALL of the information the book gives for this problem. in one of my answers x must be >= 0, in the other x must be <= 0 in order for the function to be defined so they are not the same. In the book solution x must be <= 0, but it is a different function than the one I keep coming up with. I don't see how they get that solution.

 

[mmm4444bot]

Graph the three functions that you have. Then you will know which one (if any) matches.

 

[daon]

The quatinty -(y-1)^2 is always negative. In your first attempt you forgot an absolute value.

given that x+(y-1)^2 = 0, you must have that x is always nonpositive. This is also not a function of x.

edit: whoops. just read the whole question. your asnwer will have a "plus or minus" in it. you must decide which corresponds to the bottom.