## anonymous 5 years ago how do I graph y^2+4x=0 using the (x-h)^2=4p(y-k)

1. anonymous

What do you mean exactly? Are you asking how you can put the first equation into the form you've seen for the second?

2. anonymous

yes i need to graph it and find the focus the directrix and p

3. anonymous

i would really appreciate the help I'm new haha

4. anonymous

Okay, just a sec...doing multiple things at once ;)

5. anonymous

I can be patient

6. anonymous

Okay...

7. anonymous

Your first and second equations are both parabolas, but they're different. The second equation will look like ones you're used to, whereas the first will open up differently. $y^2+4x=0 \rightarrow y^2-4x$(i.e. subtracting 4x from both sides) puts your parabola in a form where you can read off things like focus and directrix. I'll continue in a sec.

8. anonymous

So, I made a mistake...great! It should be, $y^2+4x=0 \rightarrow y^2=-4x$

9. anonymous

Now, do me a favor and go to this website: www.wolframalpha.com and in the box at the beginning, type in: y^2+4x=0 Look at the graph and then come back. I'll continue with the focus and directrix.

10. anonymous

:) ty

11. anonymous

i think its suppose to be x= -y^2/4(1)

12. anonymous

because the focus is (1,0) and the directirx is (-1,0)

13. anonymous

but thank u for your help soo much it was lovely ;)

14. anonymous

The parabola you have is one that opens on its side. The form of this kind of parabola is$(y-k)^2=4a(x-h)$where a is your focus and (h,k) is the vertex. In your case, both h and k are zero, and if you compare the form with your equation, you'll see,$y^2=4ax \rightarrow a=-1$ so your focus is at (-1,0) and directrix (1,0)

15. anonymous

hmm you have given me a lot to think about you have the opposite opinion of someone else i asked hmm ok ty :)

16. anonymous

You're welcome. Your equation above, x=-y^2/4(1) is the same as what I gave you...just looks different.

17. anonymous

thank you so much i really appreciate it :D