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 20:04:46 GMT
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Viewed:
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395 times
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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.
>
> Here's what you have to do to find the formula for a parabola, given three
> points:
>
> y = ax^2 + bx + c
>
> plug in your three points:
> 180 = a(30)^2 + b(30) + c
> 180 = a(-30)^2 + b(-30) + c
> 100 = a(14)^2 + b(14) + c
>
> You have three equations with three unknowns. From the first two, you can
> see that b has to be 0, so we're left with:
>
> 180 = 900a + c
> 100 = 196a + c
>
> 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
>
> Jeff J
> (this time I double checked everything before clicking 'post' like a good
> geek should)
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
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Message has 2 Replies:
 | | 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)
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Message is in Reply To:
| | 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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