3x + y^2 +8y + 4 = 0?

Question

I don't think I can ever figure this out. Any takers! Here are the instructions. "find the vertex, focus, and directrix of the the parabola given by each equation. Sketch the graph." I'll take any help I can get, thanks.

Thanks Diane that does help but what about the rest of the problem, like the focus, and directrix any ideas on that stuff.

How would I check this answer on a scientific calculator. I have an older one the TI-82?

Answer 1

First off, you should be able to recognize that this is a parabola (because of x and y, one of them is at most linear and the other is at most quadratic) and that this parabola will open "sideways" - i.e. along the x-axis, not the y-axis (because it's the y that's squared). That gives you your start.

Now: the standard form for parabolas is

y - k = a(x - h)^2 for "up/down" parabolas

and

x - h = a(y - k)^2 for "left/right" parabolas.

The vertex will be (h, k). The a will determine if it opens in the positive or negative direction and how steep it is. We need to get your equation into this form. We'll do it by completing the square.

Start by separating things out:

3x + 4 = -y^2 - 8y

Now complete the square:

3x + 4 = -(y^2 + 8y)

3x + 4 - 16 = -(y^2 + 8y + 16)

(I'm assuming if you're working on this, you already know how to complete the square. If not, ask about it.)

3x - 12 = -(y + 4)^2

Now divide out:

x - 4 = -(1/3)(y + 4)^2

So your vertex will be (4,- 4). The parabola will open in the negative x direction, and will be fairly shallow.

Does that help?

Answer 2

Diane's answer is correct.
Try it on Graphic Calculator, such as TI 84, you can double check this answer.

Answer 3

did you try a graphing calculator? im in reg math analysis, so we dont learn about that stuff, sorry!