Question

(y^2)+2y-2x-7=0

Answer 1

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

Answer 2

is that u were looking for?

Answer 3

She has posted so many questions about graphing I don't even know what is the question anymore.

Answer 4

im just not sure how to find the points for them

Answer 5

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?

Answer 6

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

Answer 7

check this one out

Answer 8

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.

Answer 9

did u see that?

Answer 10

too much flipping....:D

Answer 11

ahh im so confused ha

Answer 12

which part r u confused of?

Answer 13

how to get the point and how to determine the height or width of it

Answer 14

i mean i know how to get the vertex but I dont know how to figure out how much it opens up

Answer 15

okay..for that put x=0 that will give u the point where it touches the y axis and vice versa...got it?

Answer 16

do you mean to put x=0 when its like this y=1/2x^2=4x=11 ?

Answer 17

(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?

Answer 18

thank you