anonymous
  • anonymous
how do I graph y^2+4x=0 using the (x-h)^2=4p(y-k)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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?
anonymous
  • anonymous
yes i need to graph it and find the focus the directrix and p
anonymous
  • anonymous
i would really appreciate the help I'm new haha

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anonymous
  • anonymous
Okay, just a sec...doing multiple things at once ;)
anonymous
  • anonymous
I can be patient
anonymous
  • anonymous
Okay...
anonymous
  • 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.
anonymous
  • anonymous
So, I made a mistake...great! It should be, \[y^2+4x=0 \rightarrow y^2=-4x\]
anonymous
  • 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.
anonymous
  • anonymous
:) ty
anonymous
  • anonymous
i think its suppose to be x= -y^2/4(1)
anonymous
  • anonymous
because the focus is (1,0) and the directirx is (-1,0)
anonymous
  • anonymous
but thank u for your help soo much it was lovely ;)
anonymous
  • 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)
anonymous
  • anonymous
hmm you have given me a lot to think about you have the opposite opinion of someone else i asked hmm ok ty :)
anonymous
  • anonymous
You're welcome. Your equation above, x=-y^2/4(1) is the same as what I gave you...just looks different.
anonymous
  • anonymous
thank you so much i really appreciate it :D

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