 | | Re: Yet another math problem
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| (...) Hmm. Well, now that I think of it, the vertex could be x=0 with y as an unknown. Does that change anything? Dave! (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Yes, it does. If you just have the last four points to work with, then you definitely have an upward-turned parabola, symmetrical about the y-axis. Here's what you have to do to find the formula for a parabola, given three points: y = ax^2 + (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Sorry, but this doesn't go through 0,0. Solving x=0, y=0 gives 0=855/11, which is clearly not true, so this isn't the equation he's looking for either. Adrian -- www.brickfrenzy.com (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Oops. Hang on there Adrian, read for content. The five points as given in the initial problem (including 0,0) will not solve to a simple parabolic equation. Eliminating 0,0 does give us the solution specified above. Bleh. Adrian (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Hey, that's super! And I finally found an online reference at the U of Georgia site, so I can probably handle these in the future. Thanks for the clear explanation and the solution. Dave! (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) SNIP (...) Per Dave's last message, y is unknown (can't be 0), which Jeff has solved to 855/11. -Rob. (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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