[MOVED] Find eqn of parabola through (-2,3), (0,3), (1,9)

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summergrl

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Feb 21, 2007
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Find an equation for the conic that satisfies the given conditions.

Parabola, vertical axis, passing through (-2,3), (0,3), and (1,9)
 
Use the general form of a quadratic and solve for the unknowns.

\(\displaystyle \L\\a(-2)^{2}+b(-2)+c=3\\a(0)^{2}+b(0)+c=3\\a(1)^{2}+b(1)+c=9\)
 
summergrl said:
In previous problems, we had to find the vertex, focus and directrix first
Click to expand...
You found this from the random points...? (When given three random points, as opposed to "nice" points like vertices, etc, one customarily uses a system of equations, is why I ask. But if you've been taught and shown a different method, then we can work with that....)

Please reply showing how far you have gotten in working through this process. When you reply, please provide the formulas they gave you to work with.

Thank you.

Eliz.
 
look at your first two points ... (-2,3) and (0,3) ... they have the same y-value, don't they?

using the fact that a parabola has an axis of symmetry, what does this tell you about the x-value of the vertex?
 
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