I have looked all over, and I have found different things here and there for figuring pretty much everything but that.

What I have is an x chord and a y chord, but I need to find the vertex.

This isn't for school or anything, it is for an algorithm for a program I am writing.

I never got over college algebra even when I was in college and I'm certain I have learned this before, but it has been literally years since then.

So, how do I find the vertex if I have an x chord and a y chord and that is all? And what mucks it up even more is I need a reliable means of retrieving the x and y chords back again.

Realistically I can either use one of my points as x intercept and one y intercept, or I can use one as x intercept and the other the vertex. But in either case I need a reliable means of moving back and forth from the points given to a said point and back.

I have two points and I need to save one, and be able to reliably get the two original points back at will from that saved single point.

As you can see this is not an easy task and hence my brain pain. lol

Any assistance would be greatly appreciated.

Please allow me to broaden it a bit further.

I have a list of sorts which resembles a parabola.

What I really have is two points within a "shaded" region which is a parabola or bell curve.

I need to some how store these two points via some equation with a single point, and with that single point be able to retrieve the original 2 points.

I know there is a method of doing this, but the math to do it is over my head. I'm plenty smart, I just haven't learned it yet.

If it helps at all here is a python code snipet demonstrating the values which would equate to whole number chords on a graph. My program will take two of those values, equate them to chords on a graph, figure a singular value to uniquely represent those two chords, save that value, and then later on retrieve those original chords(and values) for later calculation.

def main():
    clist = []

    for x in range(0, 37):

    for a in range(0, 10):
        for s in range(0, 10):
            for d in range(0, 10):
                for f in range(0, 10):
                    if clist[a+s+d+f] == "": 
                        clist[a+s+d+f] = str(a) + str(s) + str(d) + str(f) 
                        clist[a+s+d+f] = clist[a+s+d+f] + "," + str(a) + str(s) + str(d) + str(f)

    count = 0
    for x in range(0, 37):
        datalst = clist[x].split(',')
        datacnt = len(datalst)
        count += datacnt
        print(str(x) + ", " + str(datacnt) + ", " + clist[x])

The results of which are: http://pastebin.com/SjtBMUqa (formatting bad, but you get the picture)

  • 6
    $\begingroup$ Two points do not determine a parabola. Rotate the axes and translate and rescale so that the two points are at $x=1, y=0$ and $x=-1, y=0$ - any parabola of the form $y=c(x^2-1)$ will pass through those two points. Reverse the rescaling, translation and rotation and you have a one-parameter family of parabolas which pass through the two points. $\endgroup$ – Mark Bennet Sep 1 '13 at 20:36
  • 2
    $\begingroup$ Why do you think two points will determine a parabola uniquely? $\endgroup$ – Macavity Sep 1 '13 at 20:36
  • $\begingroup$ Mark - would you dummy that down a bit. :) and Macavity - I elaborated further above. Thank you both. $\endgroup$ – texasman1979 Sep 1 '13 at 20:47
  • $\begingroup$ forgot to add the @mark $\endgroup$ – texasman1979 Sep 1 '13 at 21:10
  • $\begingroup$ I am now not sure if you are looking really for a parabola. If a Bell curve is fine, then it needs only two parameters - i.e like a point $(N, \sigma)$ to be chosen which can uniquely determine the graph. However it really doesn't have an $x$ intercept, unless you fix on a $3 \sigma$ limit or some such thing, so not sure if that helps. $\endgroup$ – Macavity Sep 1 '13 at 21:48

Usually two points do not determine a parabola because the space of parabolas has a typical point $f(x)=Ax^2+Bx+C$ and as you can count there are three parameters to fix: $A,B,C$. However, if we are given the very special point $(h,k)$ called the vertex and we are given another point $(a,b)$ such that $k \neq b$ and $h \neq a$ then these two points fix a parabola as follows: first the fact $(h,k)$ is the vertex indicates that $$ f(x) = A(x-h)^2+k $$ where $A$ is a constant. Continuing, $f(a)=b$ shows $A(a-h)^2+k=b$ so $A=\frac{b-k}{(a-h)^2}$. In summary, the pair of points $(h,k)$ and $(a,b)$ fix the parabola: $$ y=f(x) = \frac{b-k}{(a-h)^2}(x-h)^2+k. $$ All of this said, you should realize I'm cheating. These are not just any old pair of points on the parabola. No, $(h,k)$ is the vertex which means that it is at a horizontal tangent and so implicit within the statement $(h,k)$ is the vertex is both the condition the point be on the graph $k=Ah^2+Bh+C$ and the condition that the vertex have a horizontal tangent $0=2Ah+B$.

  • $\begingroup$ I added to my original post. Please forgive me, but the years have been long since I could readily understand what you just said. Could you please elaborate further? Thx. $\endgroup$ – texasman1979 Sep 1 '13 at 21:13
  • $\begingroup$ @texasman1979 Sure, if I tell you $(1,0)$ is the vertex and $(0,1)$ is a point on the parabola then the parabola has equation $y=(x-1)^2$ end of story. However, if I just tell you $(0,1)$ and $(1,0)$ are points on a parabola then there is a whole family of parabolas which fit the criterion. In my example, $f(x) = Ax^2-(1+A)x+1$ will go through $(0,1)$ and $(1,0)$ no matter my choice for $A$. For example, $A=1$ gives us $y=x^2-2x+1$ whereas $A=-1$ gives $y=-x^2+1$. Both of these contain $(0,1)$ and $(1,0)$ in their graph. $A>0$ opens up, $A<0$ opens down and $A=0$ gives a line. $\endgroup$ – James S. Cook Sep 1 '13 at 21:40
  • $\begingroup$ Say points (1,3) and (3,1) are my two points, and say I made each point the vertex of the other. Similar to inverse parabolas. So (1,3) would be vertex with point (3,1) and another where (3,1) is the vertex with point (1,3). How might I write an equation representing both relatively equal to each other and would there be a point or some other information I can use to solve both, if it is understood there correlation? Is that the A spoke of in the above? $\endgroup$ – texasman1979 Sep 1 '13 at 22:29
  • $\begingroup$ By doing that, the two parabolas will only have 2 points in common. I really am trying to resolve this, and I understand whats going on, I just have very little idea how to actually write the equation. Then I have the great chore of translating into programming language. lol $\endgroup$ – texasman1979 Sep 1 '13 at 22:38
  • $\begingroup$ @texasman1979 Wait a second, you want to allow for arbitrary parabolas? Not just those which open vertically. What about rotated parabolas? I assumed the parabolas are vertical. $\endgroup$ – James S. Cook Sep 2 '13 at 0:21

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