Find an equation for the parabola that has its vertex at the origin and satisfies the given conditions. focus on the negative y-axis, 6 units away from the directrix.

y = - x²/12

Step-by-step explanation:

The axis of symmetry passes through both the focus and vertex, and both of those are on the y-axis.  So, the y- axis *is* the axis of symmetry. The equation has the form y = ax^2 and since the focus is *below* the vertex, the parabola opens downward and the coefficient a is negative.

The vertex is halfway between the focus and the nearest point on the directrix, so the focal length (distance from vertex to focus is) 6/2 = 3. The signed focal distance is f = -3, since it's below the vertex.

A point (x,y) is on the parabola if it's equidistant from the line y=3 and the focus F=(0,-3).  

|y - 3| = √[(x - 0)² + (y + 3)²]   distance to directrix = distance to F

(y - 3)² = x² + (y + 3)²

y² - 6x + 9 = x² + y² + 6y + 9

-12y = x²

y = - x²/12

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )