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Question 813137: Find the vertex, focus, and directrix of the parabola.
Found 2 solutions by lwsshak3, TimothyLamb:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertex, focus, and directrix of the parabola.
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1/2)x^2+2x=2y+4
complete the square:
(1/2)(x^2+4x+4)=2y+4+2
(1/2)(x+2)^2=2y+6
(x+2)^2=4y+12
(x+2)^2=4(y+3)
This is a parabola that opens up.
Its basic form of equation: (x-h)^2=4p(y-k)^2,(h,k)=(x,y) coordinates of the vertex.
vertex: (-2,-3)
axis of symmetry: x=-2
4p=4
p=1
focus:(-2,-2)(p-distance above vertex on the axis of symmetry)
directrix:y=-4(p-distance below vertex on the axis of symmetry)
Answer by TimothyLamb(4379)  (Show Source):
You can put this solution on YOUR website! (1/2)x^2 + 2x = 2y + 4
(1/2)x^2 + 2x - 4 = 2y
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y = (1/4)x^2 + x - 2
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the above quadratic equation is in standard form, with a=1/4=0.25, b=1, and c=-2
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to solve the quadratic equation, plug this:
0.25 1 -2
into this: https://sooeet.com/math/quadratic-equation-solver.php
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Answer 1:
the vertex of the parabola is a minimum point at: ( -2, -3 )
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Answer 2:
the parabola has a focus at: ( -2, -2 )
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Answer 3:
the parabola has a directrix at: y = -4
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