I know it's a degenerate case because it equals 0, but it's a parabola so I'm not sure what to call it.
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$\displaystyle y^2+8x=0 \Leftrightarrow y^2=-8x$ (which means their are no real solutions, only complex) Maybe you can consult wolphram alpha to graph $\displaystyle y^2=-8x$. You can just call it parabola.
Originally Posted by explodingtoenails I know it's a degenerate case because it equals 0, but it's a parabola so I'm not sure what to call it. This conic section is a parabola with the general equation $\displaystyle (y-h)^2=4p(x-k)$ That means your equation describes a parabola opening to the left, the vertex at V(0, 0) and the p = -2 and $\displaystyle x \le 0$.
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