anonymous
  • anonymous
i need to write the vertex form of the following parabola: 1/2(y+4)=(x-7)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
anonymous
  • anonymous
put x-7=0 and y+4=0 solve and you get the vertex.
anonymous
  • anonymous
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anonymous
  • anonymous
\[{1\over2}(y+4)=(x-7)^2\]\[y+4=2(x-7)^2\]\[y=2(x-7)^2-4\]
anonymous
  • anonymous
what you get same thing from my calculation x=7 y=-4 vertex (7,-4)
anonymous
  • anonymous
yes but why would you do that when the vertex is given in the equation..
anonymous
  • anonymous
the form it was in:\[{1\over a}(y-k)=(x-h)^2\]where \((h,k)\) is the vertex the form we need is the standard form:\[y=a(x-h)^2+k\]you literally just rearrange it :)
anonymous
  • anonymous
|dw:1378774874608:dw|
anonymous
  • anonymous
correction (y-k)^2 inn the parabolas in the first row.
anonymous
  • anonymous
bro you're sidetracking....the question asks for the vertex form of a standard form equation...and it's been answered, don't confuse the asker
anonymous
  • anonymous
ok one stupid question yummydum, how did you cancel out the 1/2 and get the 2 on the other side
anonymous
  • anonymous
you don't cancel, you "multiply by the reciprocal" basically saying that since 2 is being divided, we just multiply by it on both sides:\[{1\over2}(y+4)=(x-7)^2\]\[{\cancel2\over\cancel1}\times{\cancel1\over\cancel2}(y+4)={2\over1}\times(x-7)^2\]\[y+4=2(x-7)^2\]
anonymous
  • anonymous
Yup that's what i did. Thanks!

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