SOLUTION: Find the vertex and directrix of 2x^2+4y-8=0

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Question 1167982: Find the vertex and directrix of
2x^2+4y-8=0
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B2y-4=0
2%28y-2%29=-x%5E2
-x%5E2=2%28y-2%29
vertex is at (0,2).


4p=2
p=1%2F2

Since parabola opens downward with vertex maximum, directrix is at y=2%261%2F2.