Answer 1

first we draw a picture to get our bearings

Answer 2

okay

Answer 3

the vertex of our parabola is going to be the midpoint between the d and f; in this case its (0,0)

Answer 4

the parabola opens up towards the focus soo

Answer 5

this is of the form:

y^2 = 4ax ; where a half the distance from d to f

Answer 6

the distance from -3 to 3 = 6; so a = 6/2 = 3

Answer 7

y^2 = 4(3)x

y^2 = 12x

Answer 8

so the answer is y = 1/12 x^2?

Answer 9

http://www.wolframalpha.com/input/?i=y%5E2%3D12x

no, the equation is: y^2 = 12x

Answer 10

okay but its not my choices
so the one similiar to that is
x = 1/12y^2

Answer 11

similar aint gonna cut it :) let me see the options so that I can determine a suitable choice

Answer 12

1
x = --- y^2 is adequate as long as the y^2 aint in a denominator
12

Answer 13

Okay
x = 1/12y^2
x = - 1/12 y^2
y = -1/12 x^2

x= 1/12y^2

Answer 14

then yes :) x = 1/12 y^2

Answer 15

I have two more can i do it myself and you tell me if its correct or not?

Answer 16

i can dbl chk your results yes

Answer 17

Write the equation of a parabola with a vertex at (0, 0) and a focus at (-4, 0).

y= -1/16 x^2 ?