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

1 Answer
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(y-k)^2#
#x=-y^2+2y=-(y^2-2y )#
so #x=-(y^2-2y+1-1)=-(y^2-2y+1)+1#. Therefore,
#(x-1)=-(y-1)^2#

The vertex is at #(h, k)=(1, 1)#

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 #(1, 1)# and go #+-1# in the #y# direction (vertically), the we'll go #a=-1# in the x direction (horizontally) This gives us 2 additional points #(0, 0)# and #(0, 2)#.

That's enough to sketch the graph:

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