Which is the standard form of the equation of a parabola with a focus of (0, -3) and its vertex at the

it’s A

Step-by-step explanation:

1:This is a parabola by the looks of the equation and the fact that it's a satellite dish. The vertex of the equation is (-6,0) and the focus is 2/4, 1/2. 1/2 plus -6 is -5.5. the focus is (-5.5,0).

This answer is only true if the actual equation is y^2 not 2y 
2:y=-(x+3)^2/8+1

3:This is the answer(–1.25, –3),

4:
answer is: x^2=-12y

ANSWER

[tex]y = - \frac{1}{12} {x}^{2}[/tex]

EXPLANATION

The parabola has its vertex at the origin and its focus is at (0,-3).

This implies that, the parabola opens downwards.

The equation of such parabola is of the form:

[tex]{x }^{2} = - 4py[/tex]

p is the focal length. The distance from the focus to the vertex.

p=0--3=3

[tex]{x }^{2} = - 4(3)y[/tex]

[tex]{x }^{2} = - 12y[/tex]

Or

[tex]y = - \frac{1}{12} {x}^{2}[/tex]

the answer on edg is B for anyone using that!!!

the equation of the parabola is y2=12x

explanation:

the vertex is midway between the focus ad the directrix

so, the vertex is (0,0)

let p=(x,y) be a point on the parabola.

the distance of p from the directrix is equal to the distance from the focus.

x+3=√(x−3)2+(y−0)2

(x+3)2=(x−3)2+y2

x2+6x+9=x2−6x+9+y2

y2=12x

graph{y^2=12x [-10, 10, -5, 5]}