I need to find the equation of the cubic curve that passes through four points $(1,-126)$, $(8,840)$, $(-6,280)$ and $(-11,270)$.

Using the form $y = ax^3 + bx^2 +cx +d$, I produced $4$ equations:

From $(-11,270):-1331a+121b-11c+d=270,$

From $(-6,280):-216a+36b-6c+d=280,$

From $(8,840): 512a+64b+8c+d=840,$

From $(1,-126): a+b+c+d=-126,$

I have attempted to solve these by elimination, subtracting each equation by the one below (excluding the bottom), but the answer doesn't satisfy all points. With slight changes to my method, it has produced several different answers that never intersect all $4$ points.

Am I just making a mistake in subtraction, or do I need to use a different method? Can a simultaneous equation have no solutions? Any help would be very much appreciated.

  • $\begingroup$ Is it $$y=ax^3+bx^2+cx+d$$? $\endgroup$ – Dr. Sonnhard Graubner Jun 12 '19 at 11:13
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    $\begingroup$ @Dr.SonnhardGraubner Some times (like this time), single dollar signs would look a lot better than double dollar signs. $\endgroup$ – Arthur Jun 12 '19 at 11:16
  • $\begingroup$ It is $a=1,b=11,c=-34,d=-104$ $\endgroup$ – Dr. Sonnhard Graubner Jun 12 '19 at 11:24
  • $\begingroup$ Thank you, that works. Tried it again knowing the answer and got it. $\endgroup$ – Connor Muir Jun 13 '19 at 0:31

Polynomial interpolation "always works" (if the abscissas are different).

  • the determinant of the system is in Vandermonde form and is nonzero;

  • an explicit solution is given by the Lagrangian formulas.

Try again using elimination, which is the most convenient by hand (eliminate $d$ first).

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  • $\begingroup$ Thanks for the alternative suggestions. $\endgroup$ – Connor Muir Jun 13 '19 at 0:33

Sometimes it's hard to keep track of the signs correctly when subtracting. You might make mistakes.

An alternative to subtracting equations is adding them. You can change the sign of one of the equations in order to obtain the cancellation that subtraction would have given you.

In this problem, the last equation is equivalent to $$ -a - b - c - d = 126.$$

You can add this same equation to each of the first three equations. You aren't restricted to only combining equations that were next to each other. The results from these operations on the first three equations would be: \begin{align} -1332a+120b-12c &= 396, \\ -217a+35b-7c &= 406, \\ 511a+63b+7c &= 966. \\ \end{align}

Now obviously you can add the last two equations to get an equation without $c.$ But the first equation would be easier to deal with if the coefficient of $c$ were $1$ in all three places, wouldn't it? So try dividing the first equation by $12$ and each of the other two by $7.$

If it still doesn't work you might try showing every step of at least one attempt in the question so someone will have a chance of seeing what is going wrong.

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  • $\begingroup$ Thank you very much. Tried this and got it eventually. $\endgroup$ – Connor Muir Jun 13 '19 at 0:33

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