# How do you graph x=2y-y^2?

Mar 9, 2015

If you have studied graphing parabolas, then you want to use the standard method for them.

Because the $y$ is square, not the $x$, this parabola opens horizontally.
(There's more than one way to proceed. I'll use one of them.)
Get the equation in standard form: $x - h = a {\left(y - k\right)}^{2}$
$x = - {y}^{2} + 2 y = - \left({y}^{2} - 2 y\right)$
so $x = - \left({y}^{2} - 2 y + 1 - 1\right) = - \left({y}^{2} - 2 y + 1\right) + 1$. Therefore,
$\left(x - 1\right) = - {\left(y - 1\right)}^{2}$

The vertex is at $\left(h , k\right) = \left(1 , 1\right)$

The coefficient of the square term is negative, so the parabola opens in the negative direction. For a horizontal parabola, this means it opens to the left.

The axis of symmetry is the line $y = 1$.
And if we start at the vertex $\left(1 , 1\right)$ and go $\pm 1$ in the $y$ direction (vertically), the we'll go $a = - 1$ in the x direction (horizontally) This gives us 2 additional points $\left(0 , 0\right)$ and $\left(0 , 2\right)$.

That's enough to sketch the graph:

graph{x=2y-y^2 [-4.933, 4.932, -2.466, 2.467]}