## Sketching this curve

1. The problem statement, all variables and given/known data

I need to make a sketch of this function:

4x = 4y - y^2

2. Relevant equations

3. The attempt at a solution

So I see that it's a kind of parabola in terms of y.. so I try to make it into parabola form:

x = y - (1/4)y^2
= (-1/4)(y^2-4y+4-4+0)
= (-1/4)((y-2)^2-4))
= (-1/4)(y-2)^2+1

Now I try to write it in terms of x:
x-1 = (-1/4)(y-2)^2
-4x+4 = (y-2)^2
+/-sqrt(-4x+4) = y-2
y = +/-sqrt(-4x+4)+2

But these curves look like they are different?
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 your functions should look like this: Attached Thumbnails

Mentor
 Quote by zeion 1. The problem statement, all variables and given/known data I need to make a sketch of this function: 4x = 4y - y^2 2. Relevant equations 3. The attempt at a solution So I see that it's a kind of parabola in terms of y.. so I try to make it into parabola form: x = y - (1/4)y^2 = (-1/4)(y^2-4y+4-4+0) = (-1/4)((y-2)^2-4)) = (-1/4)(y-2)^2+1 Now I try to write it in terms of x: x-1 = (-1/4)(y-2)^2
Leaving it in the form above is helpful, as you can tell that the graph is similar to the graph of x = -y^2. This is a parabola whose axis of symmetry is horizontal, and that opens to the left. Your parabola can be thought of as the translation to the right by 1 unit and up 2 units of the graph of x = -y^2. This puts the vertex at (1, 2).
 Quote by zeion -4x+4 = (y-2)^2 +/-sqrt(-4x+4) = y-2 y = +/-sqrt(-4x+4)+2 But these curves look like they are different?
Yes. What you are getting by solving for y are equations for the upper and lower halves of the parabola. The pos. square root gives the upper half, and the neg. sq. root gives the lower half.