SOLUTION: Find the vertex,focus,directrix,and axis of symmetry in the given equation (x+3)^2= -24y+60

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Question 1041228: Find the vertex,focus,directrix,and axis of symmetry in the given equation
(x+3)^2= -24y+60
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
(x+3)^2= -24y+60
(x-h)^2=4p(y-k)
-24y+60=-24(y-2.5)
(x+3)^2=4*(-6)(y-2.5)
focus is (h,k+p) or (-3,-2.5-6) or (-3,-8.5)
directrix is y=(k-p)=2.5-(-6)=8.5
can rewrite as -24y=(x+3)^2-60
divide both sides by -24 and y=(-1/24)(x+3)^2+2.5. The vertex is (-3,2.5)
The axis of symmetry is at x= -3
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