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.Com
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)   (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 = ± - 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

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