SOLUTION: what is the answer to this equation?i'm looking for it's vertex and everything needed to draw it's parabola.(y+3)^2 = -6(x-6). marilag.michelle@yahoo.com

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: what is the answer to this equation?i'm looking for it's vertex and everything needed to draw it's parabola.(y+3)^2 = -6(x-6). marilag.michelle@yahoo.com      Log On


   

Question 409366: what is the answer to this equation?i'm looking for it's vertex and everything needed to draw it's parabola.(y+3)^2 = -6(x-6).
marilag.michelle@yahoo.com
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

sketching parabola: Format: (y-k)^2 = 4p(x-h)

(y+3)^2 = -6(x-6)

I. Vertex is Pt(6,-3)

II. Opens left.  Line of symmetry is y = -3

III. x = 0, y = ±sqrt%2836%29 - 3 |  Pt(0,3) and Pt(0,-9) the y-intercepts

IV. 4p = -6, p = -3/2     | Focus is Pt( (6 -3/2), -3)= Pt(4.5,-3)

V. Directrix is x = 6 + 3/2 =  7.5