Originally Posted by
HallsofIvy Your question is not at all clear. You could "sketch" the graph by calculating several points: if x= 0, $\displaystyle 2y^2+ y= y(2y+1)= 0$ so y= 0 or -1/2. $\displaystyle (0, 0)$ and $\displaystyle (0, -\frac{1}{2})$ are points on the graph. Taking y= 0, we get, in exactly the same way, $\displaystyle (0,0)$ which we had before, $\displaystyle (0,\frac{1}{2})$. Get enough points that way and you can sketch the graph.
But I suspect that you are intended to complete the square to see what kind of figure this is:
If you divide through by 2 you get $\displaystyle x^2+ \frac{1}{2}x+ y^2+ \frac{1}{2}y= 0$. Presumably you know that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$. Comparing that to $\displaystyle x^2+ \frac{1}{2}x$, they match if a= 1/4 and we can make this a "perfect square by adding $\displaystyle \left(1/4\right)^2= 1/16$. Doing that for both x and y, $\displaystyle x^2+ \frac{1}{2}x+ \frac{1}{4}+ y^2+ \frac{1}{2}y+ \frac{1}{4}= \frac{1}{4}+ \frac{1}{4}$ or $\displaystyle (x+\frac{1}{2})^2+ (y+\frac{1}{2})^2= \frac{1}{2}$. What kind of figure is that?