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Subject: 
Re: Yet another math problem
Newsgroups: 
lugnet.off-topic.geek
Date: 
Mon, 11 Nov 2002 18:54:59 GMT
Viewed: 
368 times
  
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
My bad. :(

Jeff J



Message has 1 Reply:
  Re: Yet another math problem
 
(...) 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)

Message is in Reply To:
  Re: Yet another math problem
 
(...) 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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