Quboid Posted June 13, 2004 Share Posted June 13, 2004 If I have three points (normal x,y) on a graph, how do I calculate the formula of the curve that would run through these three points? I've sure I learnt about this stuff back at school but damned if I remember. That's what comes of not listening... :blush: Link to comment Share on other sites More sharing options...
ZTrang Posted June 13, 2004 Share Posted June 13, 2004 If I have three points (normal x,y) on a graph, how do I calculate the formula of the curve that would run through these three points?I've sure I learnt about this stuff back at school but damned if I remember. That's what comes of not listening... :blush: What type of curve? You need to be more specific. I assume you mean a quadratic curve (i.e. x^2+2*x+1), but please specify before I go to the trouble explaining how to solve for one. Link to comment Share on other sites More sharing options...
Quboid Posted June 13, 2004 Author Share Posted June 13, 2004 Yes, quadratic, sorry. I'm pretty sure it's simply (x^z) - 1, where z is presently unknown, but I'm not certain. Link to comment Share on other sites More sharing options...
ZTrang Posted June 13, 2004 Share Posted June 13, 2004 Yes, quadratic, sorry.I'm pretty sure it's simply (x^z) - 1, where z is presently unknown, but I'm not certain. Well, there is a simple way to solve for an equation after you have three points. You see, the formula for any quadratic equation is ax^2+bx+c=y; you want to find a,b, and c, then. To do this, plug in your points for x and y to get three simultaneous equations, for example, using the points (1,2), (4,6), (-1,5), one would get 2=a+b+c 6=16a+4b+c 5=a-b+c Then just solve those equations using the normal methods for solving simultaneous systems of equations (I won't explain it here, as they are pretty easy and tedious; look them up if you don't know them), and then you'll have your quadratic equation. Link to comment Share on other sites More sharing options...
Quboid Posted June 13, 2004 Author Share Posted June 13, 2004 OK, I think I get it. So for (x,y), one of the equations is: y = (x^2)a + (x^1)b + (x^0)c or y = (x^2)a + xb + c Right? Once I get the forumulas, it's no problem (but tedious, yes) to solve. Link to comment Share on other sites More sharing options...
ZTrang Posted June 13, 2004 Share Posted June 13, 2004 Exactly. You could always use a calculator program to do it for you (I think some such programs are made for the TI-83 Plus), but they use the method I have described to get the equation. Link to comment Share on other sites More sharing options...
Louisville Slugger Posted June 13, 2004 Share Posted June 13, 2004 (edited) Hehe ok I'll give you a cool example. Let?s say I have the costs of a 10", 12" and a 14" pizza, which would be $5.95, $6.95, and $8.50 respectively. The length of the pizza would be the (x) variable and the price would be the y variable.i>---------------------b> (10", $5.95) (12", $6.95) (14", $8.50)---------------------b>I would subtract the $6.95 from the $5.95 to get a difference of $1.00, then subtract $8.50 from $6.95 and get a difference of $1.55. Then you would subtract $1.55 from $1.00 and get a difference of $.55. Therefore we now know it's a 2nd degree polynomial so the form of the equation would be y = ax^2 + bx + c.i>---------------------b>Now let's say I wanted to find what the price of a 16" pizza would be. I would take those three points and plug them into the y = ax^2 + bx + c form of the equation so that the prices of the pizza are going from greatest to least.i>---------------------b>1.) $8.50 = 196a + 14b + c 2.) $6.95 = 144a + 12b + c 3.) $5.95 = 100a +10b + ci>---------------------b> Now, the successive subtraction begins. Subtract equation numbers 1 and 2. $1.55 = 52a + 2b (the c term cancels out) Subtract equation numbers 2 and 3. $1.00 = 44a + 2b (the c term cancels out) Now subtract these two equations . . . $.55 = 8a Divide both sides by 8 and (a) is equal to .06875. Plug (a) back into one of the equations with (a) and (b) in it to solve for (b). $1.00 = 44(.06875) + 2b $1.00 = 3.025 + 2b Subtract 3.025 from both sides -2.025 = 2b Divided by 2 on both sides b is equal to -1.0125 Now take an equation with (a), (b), and ? in it and solve for ?. $5.95 = 100(.06875) + 10(-1.0125) + c $5.95 = 6.875 + -10.125 + c $5.95 = -3.25 + c Add 3.25 on both sides c is equal to 9.2--------------------- Now once you have (a), (b), and ?, plug them into the formula of a quadratic: y = ax^2 + bx + c. y = .06875(x)^2 -1.0125(x) + 9.2 Now since I want to find how much a 16" pizza would cost me I would plug in 16" for x, and I'll show it using the Euler notation. f(16") = .06875(16")^2 -1.0125(16") + 9.2 = 17.6 - 16.2 + 9.2 = $10.60 for a 16" p:Dza :D Hmmm I am hungry now . . . .<To check this using your TI-83 if you have one, go to the STAT menu, 1: (Where it says EDIT), enter in the X variables for L1, and the Y variables for L2. Then go to STAT -- CALC, 5: (QuadReg), and type 2nd key L1, 2nd key, L2, and go to VARS -- Y-VARS, 1: (Function) and hit enter. Once you have hit enter again your calculator should display the equation and all the answers to the (a), (b) and ? variables.s. Edited June 13, 2004 by Louisville Slugger Link to comment Share on other sites More sharing options...
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