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 20:14:48 GMT
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Viewed:
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521 times
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In lugnet.off-topic.geek, Adrian Drake writes:
> In lugnet.off-topic.geek, Jeff Jardine writes:
> > In lugnet.off-topic.geek, Dave Schuler 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)
> > >
> > >
> > > Hmm. Well, now that I think of it, the vertex could be x=0 with y as an
> > > unknown. Does that change anything?
> >
> >
> > 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.
SNIP
> >
> > so, a = 5/44, and c = 855/11, so the vertex intercepts the y-axis at 855/11
> > (about 77.7)
> >
> > Your solution is y = (5/44)x^2 + 855/11
>
>
> 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.
Per Dave's last message, y is unknown (can't be 0), which Jeff has solved to
855/11.
-Rob.
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Message is in Reply To:
 | | 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)
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