 | | Yet another math problem —Dave Schuler
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| | | This one should be pretty simple for you math-literate folks out there, but it's giving me a dreadful time... I have five points and am trying to define the parabola that contains them (if such exists). The points are: (0,0) (which is also the (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | | | | Re: Yet another math problem —Larry Pieniazek
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| | | | | (...) Are you looking for the defining equation of the form y = Ax2+Bx+c ?? or for something else? If the former wouldn't you just solve for "A" and "B" given that "C" is known to be zero (since you said that 0,0 is a point and is the vertex it (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | | | | Re: Yet another math problem —Jeff Jardine
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| | | | (...) Ugh - I can't believe it takes a math problem to get me to post for the first time in weeks. The formula for a parabola is: y = ax^2 + bx + c You only need three points to define a parabola, but since you've made yours symmetrical about the (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | | | | | Re: Yet another math problem —Jeff Jardine
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| | | | (...) Oops - I'd better correct myself before else does. Your parabola does NOT exist. It is NOT symmetrical. I solved it for the three points with x > 0 My bad. :( Jeff J (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | | | | | Re: Yet another math problem —Dave Schuler
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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)
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| | | | | | Re: Yet another math problem —Jeff Jardine
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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)
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| | | | | | Re: Yet another math problem —Adrian Drake
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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)
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| | | | | | | Re: Yet another math problem —Adrian Drake
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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)
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| | | | | | | Re: Yet another math problem —Rob Doucette
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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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| | | | | | Re: Yet another math problem —Dave Schuler
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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)
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