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anonymous
 5 years ago
how do I graph y^2+4x=0 using the
(xh)^2=4p(yk)
anonymous
 5 years ago
how do I graph y^2+4x=0 using the (xh)^2=4p(yk)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean exactly? Are you asking how you can put the first equation into the form you've seen for the second?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i need to graph it and find the focus the directrix and p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would really appreciate the help I'm new haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, just a sec...doing multiple things at once ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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. \[y^2+4x=0 \rightarrow y^24x\](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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, I made a mistake...great! It should be, \[y^2+4x=0 \rightarrow y^2=4x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think its suppose to be x= y^2/4(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the focus is (1,0) and the directirx is (1,0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but thank u for your help soo much it was lovely ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The parabola you have is one that opens on its side. The form of this kind of parabola is\[(yk)^2=4a(xh)\]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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm you have given me a lot to think about you have the opposite opinion of someone else i asked hmm ok ty :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome. Your equation above, x=y^2/4(1) is the same as what I gave you...just looks different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so much i really appreciate it :D
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