Solving Cubic Equations: Finding a Line Through 2 Points

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Discussion Overview

The discussion revolves around how to modify a cubic equation, specifically of the form y = x^3, so that it passes through two specified points, one of which is the origin (0,0). Participants explore various methods for achieving this, including curve fitting and algebraic manipulation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses a desire to modify the cubic equation y = x^3 to pass through two predetermined points.
  • Another participant clarifies that y = x^3 is a specific curve and suggests selecting arbitrary x-values to find corresponding points on the curve.
  • A different participant proposes using cubic spline interpolation as a method for fitting a cubic to known data points.
  • One suggestion involves solving for a coefficient 'a' in the equation a x_0^3 = y_0 to ensure the curve passes through the second point.
  • Another approach discussed involves altering the cubic equation to y = x^3 + ax, with a derived from the second point's coordinates.
  • Some participants express confusion regarding notation in the equations, mistaking 'x' for 'I' and questioning the clarity of the typesetting.

Areas of Agreement / Disagreement

Participants present multiple methods for modifying the cubic equation, indicating a lack of consensus on a single approach. There is also confusion regarding notation, but no agreement on the implications of this confusion.

Contextual Notes

Participants reference different mathematical techniques and notations, which may depend on individual interpretations and familiarity with calculus concepts. The discussion includes various approaches without resolving which is the most appropriate or effective.

Who May Find This Useful

This discussion may be of interest to students or individuals seeking to understand cubic equations, curve fitting techniques, and the manipulation of polynomial functions in mathematical contexts.

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I've taken Calculus 1, but it was a few years ago, so bear with me. I understand how to use derivitaves to find critical numbers, relative max's and min's, points of inflections, incresing, decreasing. all that good stuff.

my question is, if I have two points I want a x^3 line pass through, how can i accomplish this?
 
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programmer said:
my question is, if I have two points I want a x^3 line pass through, how can i accomplish this?
y = x3 is a curve, not a line.
I don't really get what you mean. y = x3 is a specific curve, i.e you cannot change the set of points that it passes through.
If you want to get 2 points on the curve, just choose 2 arbitrary x1, and x2 values, then from there find the corresponding y1, and y2. And you'll have 2 points that the curve passes through.
 
no, i want to MODIFY a x^3 equation to MAKE it pass through the points i already have set.
 
the first point is 0,0, and the 2nd point could be anything, how to i alter the curve to make it pass through my 2nd point?
 
You're trying to curve-fit a cubic to some known data set.

Look up cubic spline interpolation. Numerical Recipes in C has a section on it, with the algorithm coded in C.

http://www.library.cornell.edu/nr/cbookcpdf.html

If you only have two points, one of which is (0, 0) and the other is (x0, y0), just solve this equation for a:

[itex]a x_0^3 = y_0[/itex]

- Warren
 
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There are many different ways to do this. For example, as chroot said, if you want a formula of the form y= ax3, which necessarily passes through (0,0) for all a, just select a so that y0= ax3: That is [tex]y= \left(\frac{y_0}{x_0^3}\right)x^3[/tex] passes through (0,0) and (x0,y0).

Or, you could alter y= x3 to look like y= x3+ ax. In order to have y= y0 when x= x0 we must have [itex]y_0= x_0^3+ ax_0[/itex] or, solving for a, [itex]a= \frac{y_0- x_0^3}{x_0}[/itex]. That is, the graph of [itex]y= x^3+ \frac{y_0- x_0^3}{x_0}x[/itex] passes through (0,0) and (x0,y0). There are many other possiblities. The choice is essentially arbitrary unless you have other conditions to fulfill.
 
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well...i guess I've forgotten a lot of calculus...what does the I in those equations stand for? Integral?
 
programmer said:
well...i guess I've forgotten a lot of calculus...what does the I in those equations stand for? Integral?
?? What I are you talking about? Which response does this relate to?
 
nevermind...the X's in those equations look like I's...

that's retarded
 
  • #10
programmer said:
nevermind...the X's in those equations look like I's...

that's retarded

Mm... that's how all x's are typeset in every math book I've ever seen.

- Warren
 

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