Question 746203: How do you write the vertex form equation of the following parabola's:
vertex at origin, directrix: y= -(1/8)
vertex at origin, directrix: y= (1/4)
vertex at origin, focus: (0, 1/8)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do you write the vertex form equation of the following parabola's:
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vertex at origin, directrix: y= -(1/8)
axis of symmetry: y-axis or x=0
Basic form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of vertex
vertex form of equation:y=A(x-h)^2+k,,(h,k)=(x,y) coordinates of vertex
parabola opens upward:
p=1/8 (distance from vertex to directrix on the axis of symmetry)
4p=1/2
x^2=(1/2)y
vertex form of equation:y=2x^2
..
vertex at origin, directrix: y= (1/4)
axis of symmetry: y-axis or x=0
Basic form of equation: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
vertex form of equation:y=A(x-h)^2+k,,(h,k)=(x,y) coordinates of vertex
parabola opens downward:
p=1/4 (distance from vertex to directrix on the axis of symmetry)
4p=1
x^2=-y
vertex form of equation: y=-x^2
..
vertex at origin, focus: (0, 1/8)
axis of symmetry: y-axis or x=0
Basic form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of vertex
vertex form of equation:y=A(x-h)^2+k,,(h,k)=(x,y) coordinates of vertex
parabola opens upward:
p=1/8 (distance from vertex to focus on the axis of symmetry)
4p=1/2
x^2=(1/2)y
vertex form of equation: y=2x^2
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