Need help with parabola

K

kittypie

New member
Joined
Dec 20, 2013
Messages
2
I 'm currently stuck on this one problem and have no idea what to do. can you please help me?

the problem is Discuss and sketch:

2x+4y^2+8y=0

so i tried doing the problem.

2x+4y^2+8y=0

2x+(4y^2+8y)=0

2x+4y(y+2y-2+2)=0
 
kittypie said:
Discuss and sketch:

2x+4y^2+8y=0
Click to expand...
I have no idea what they might mean by "discuss". You'll want to check your textbook and your class notes; whatever they "discuss" is probably what you're expected to talk about, too.

As for sketching, it'll probably be helpful to reformat first to get the equation into a form that looks a bit more familiar.

. . . . .2x + 4y^2 + 8y = 0

. . . . .2x = -4y^2 - 8y

. . . . .x = -2y^2 - 4y

. . . . .x = -2(y^2 + 2y)

. . . . .x = -2(y^2 + 2y + 1 - 1)

. . . . .x = -2(y^2 + 2y + 1) - 2(-1)

. . . . .x = -2(y + 1)^2 + 2

Graph as usual, other than for the parabola being "sideways". ;)
 
Discuss

stapel said:
I have no idea what they might mean by "discuss". You'll want to check your textbook and your class notes; whatever they "discuss" is probably what you're expected to talk about, too.

As for sketching, it'll probably be helpful to reformat first to get the equation into a form that looks a bit more familiar.

. . . . .2x + 4y^2 + 8y = 0

. . . . .2x = -4y^2 - 8y

. . . . .x = -2y^2 - 4y

. . . . .x = -2(y^2 + 2y)

. . . . .x = -2(y^2 + 2y + 1 - 1)

. . . . .x = -2(y^2 + 2y + 1) - 2(-1)

. . . . .x = -2(y + 1)^2 + 2

Graph as usual, other than for the parabola being "sideways". ;)
Click to expand...

I think by "discuss" they might be asking for, the vertex, focus, and the equations for the directrix and axis of symmetry, all available after putting the equation in parabolic form as above and can then be used to sketch the graph ... ?
 
Last edited:
You must log in or register to reply here.
Top