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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
is that u were looking for?
She has posted so many questions about graphing I don't even know what is the question anymore.
im just not sure how to find the points for them
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?
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
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.
did u see that?
too much flipping....:D
ahh im so confused ha
which part r u confused of?
how to get the point and how to determine the height or width of it
i mean i know how to get the vertex but I dont know how to figure out how much it opens up
okay..for that put x=0 that will give u the point where it touches the y axis and vice versa...got it?
do you mean to put x=0 when its like this y=1/2x^2=4x=11 ?
(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?