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 20:11:10 GMT
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Viewed:
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385 times
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In lugnet.off-topic.geek, Jeff Jardine writes:
> 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
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!
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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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