anonymous
  • anonymous
(y^2)+2y-2x-7=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
y^2+2y= 2x+7 (y^2+2y+1)-1=2x+7 (y+1)^2=2x+8 (y+1)^2=2(x+4) which is a right handed parabola
anonymous
  • anonymous
is that u were looking for?
anonymous
  • anonymous
She has posted so many questions about graphing I don't even know what is the question anymore.

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anonymous
  • anonymous
im just not sure how to find the points for them
anonymous
  • anonymous
okay...lets try to draw it.....the above equation can be written as below (y-(1))^2=2(x-(-4)). which means its centre is (-4,-1). can u locate this point now?
anonymous
  • anonymous
yeah. i just dont know how to find it myself if that makes sense well here i have another one for practice, can you go through this one with me..... (y^2)+8y-2x+22=0
anonymous
  • anonymous
check this one out
1 Attachment
anonymous
  • anonymous
This is what I would do... switch the x and y, so that (x^2)+8x-2y+22=0, and also flip the x and y axis. So that you are essentially graphing a parabola, which is easy, and then flip the page back to normal, and you get the equation you want.
anonymous
  • anonymous
did u see that?
anonymous
  • anonymous
too much flipping....:D
anonymous
  • anonymous
ahh im so confused ha
anonymous
  • anonymous
which part r u confused of?
anonymous
  • anonymous
how to get the point and how to determine the height or width of it
anonymous
  • anonymous
i mean i know how to get the vertex but I dont know how to figure out how much it opens up
anonymous
  • anonymous
okay..for that put x=0 that will give u the point where it touches the y axis and vice versa...got it?
anonymous
  • anonymous
do you mean to put x=0 when its like this y=1/2x^2=4x=11 ?
anonymous
  • anonymous
(y+1)^2=2(x+4) is our parabola. lets put x=0 , we get (y+1)^2=2(0+4) or (y+1)^2=8 or y^2+2y-7=0 which is quadratic , solving it will give us two values of y which will give us two points where the parabola intersects the y-axis.. alright?
anonymous
  • anonymous
thank you

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