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anonymous
 5 years ago
(y^2)+2y2x7=0
anonymous
 5 years ago
(y^2)+2y2x7=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y^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
 5 years ago
Best ResponseYou've already chosen the best response.0is that u were looking for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0She has posted so many questions about graphing I don't even know what is the question anymore.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im just not sure how to find the points for them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay...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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. 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)+8y2x+22=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what I would do... switch the x and y, so that (x^2)+8x2y+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
 5 years ago
Best ResponseYou've already chosen the best response.0too much flipping....:D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh im so confused ha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which part r u confused of?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how to get the point and how to determine the height or width of it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean i know how to get the vertex but I dont know how to figure out how much it opens up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay..for that put x=0 that will give u the point where it touches the y axis and vice versa...got it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mean to put x=0 when its like this y=1/2x^2=4x=11 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(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+2y7=0 which is quadratic , solving it will give us two values of y which will give us two points where the parabola intersects the yaxis.. alright?
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