How do you Find the vertex, focus, and directrix of the parabola and how do you sketch its graph?

x + y^2 = 0

Question

How do you Find the vertex, focus, and directrix of the parabola and how do you sketch its graph?

x + y^2 = 0

anonymous · Answer

start by writing $-y^2=x$

anonymous · Answer

hmm actually i think you might want the general form as $(y-k)^2=4p(x-h)$ no matter because the vertex is still $(0,0)$

anonymous · Answer

the reason to write it as $(y-k)^2=4p(x-h)$ is that $p$ tells you the focus. 
in this case we have $y^2=-x$ and so $4p=-1$ making $p=-\frac{1}{4}$ so your focus is $(-\frac{1}{4},0)$ i.e. $\frac{1}{4}$ units to the left of the vertex

anonymous · Answer

sorry i screwed up on first post. 
it should have been $(y-k)^2=4p(x-h)$ from the start 
and your job is to force your equation to look like that, i.e. interperet 
$$y^2=-x$$ in terms of $$(y-k)^2=4p(x-h)$$

anonymous · Answer

since the focus is at $(-\frac{1}{4},0)$ to the right of the vertex, we know the directrix is on the other side, the same distance away, so it is the line $x=\frac{1}{4}$

anonymous · Answer

we can see it from the nice picture http://www.wolframalpha.com/input/?i=parabola+y^2%3D-x

anonymous · Answer

wow, another typo, $(-\frac{1}{4},0)$ is to the LEFT over the vertex, so the directrix is to the right. correct answer though