anonymous
  • anonymous
what are the equations to these diagrams? (attachment) please help will medal!!
Mathematics
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anonymous
  • anonymous
what are the equations to these diagrams? (attachment) please help will medal!!
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
PLEASE HELP!

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welshfella
  • welshfella
the first one is a quadratic equation with a negative coefficient of x^2. It has zeros of 2 and -8. and a vertex at (-3,6) can you find the equation from this info?
anonymous
  • anonymous
For the 1st diagram \[-(x+3)^2=4(y-6)\]
anonymous
  • anonymous
is this right? −(x+3)2=4(y−6) @welshfella
anonymous
  • anonymous
the ellipse equation is \[\frac{ (y-4)^2 }{ 6^2 }+\frac{ (x-3)^2 }{ 4^2 }=1\]
anonymous
  • anonymous
for the first one? @saseal
anonymous
  • anonymous
yes the first picture
anonymous
  • anonymous
what about the second? @saseal
welshfella
  • welshfella
yes that is correct for the first graph
anonymous
  • anonymous
im doing the circle now, its \[(x+6)^2+(y+2)^2=5^2\]
anonymous
  • anonymous
what about #13 and #14? then i will figure out the rest myself thank you for all ur help @saseal
anonymous
  • anonymous
\[\frac{ (x-7)^2 }{ 2^2 }-\frac{ y-2)^2 }{ 2^2}=1\] for the hyperbola
anonymous
  • anonymous
circle is #13 and hyperbola is #14
anonymous
  • anonymous
what about the one on the top right?
anonymous
  • anonymous
the oval @saseal
anonymous
  • anonymous
2nd picture?
anonymous
  • anonymous
yes
anonymous
  • anonymous
gimme a few minutes
anonymous
  • anonymous
thank you so much
anonymous
  • anonymous
First oval is \[\frac{ (x-7)^2 }{ 6^2 }+\frac{ y^2 }{ 3^2 }=1\]
anonymous
  • anonymous
if i tag you in some other ones on a different question could you help?
welshfella
  • welshfella
plz dont just give answers saseal Code of Conduct asks us to Guide the user to the answers
anonymous
  • anonymous
okies
anonymous
  • anonymous
so I can't give ya answer now, work the last one out yourself; parabola equation for that curve looks something like this :) \[(y+1)^2=4p(x-4)^2\]

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