Subject:
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Re: Yet another math problem
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 11 Nov 2002 19:12:57 GMT
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Viewed:
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370 times
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In lugnet.off-topic.geek, Jeff Jardine writes:
> In lugnet.off-topic.geek, Jeff Jardine writes:
> > >
> > > 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 vertex)
> > > (14,100)
> > > (-14,100)
> > > (30,180)
> > > (-30,180)
> >
> >
> > Your parabola does exist, a = -1/14, b = 57/7, and c=0
>
>
> 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
Hmm. Well, now that I think of it, the vertex could be x=0 with y as an
unknown. Does that change anything?
Dave!
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Message has 1 Reply:
 | | 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)
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Message is in Reply To:
| | Re: Yet another math problem
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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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