Question 370545: Hi, I'm having trouble graphing the following parabola:
It seems so simple. I have the vertex (0, 0) and I'm trying to find points and I'm getting the answer wrong. I found (32, 4) and (18, 3)...but these are apparently wrong. Please help!
Thanks for your time :)
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! having trouble graphing the following parabola:
x = 2y^2
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Solve for "y":
y^2 = x/2
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y = +/-sqrt(x/2)
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Plotting points:
Let x = 0, they y = 0
Let x = 8, then y = +2 and y = -2
Let x = 32, then y = +4 and y = -4
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Plot the points: (0,0), (8,2), (8,-2), (32,4), (32,-4)
and draw a parabola that opens to the right.
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Cheers,
Stan H.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website!
...
(0,0)
(2,1)
(2,-1)
(8,-2)
(8,+2)
..
Plot the points and draw the curve.
..
It ia a parabola opening to the right
...
m.ananth@hotmail.ca
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