How to find equation of a cubic function give four random data points?

Question

Given the data points (5.89,40) (6.95,50) (8.22,60) and (9.31,70) how do I find the equation of the cubic function with Maths B calculus techniques? (without matrix) I tried solving it simultaneously, but is stuck on how to eliminate the variables.

Answer 1

Try again with a set of simultaneous equations but you can apply a couple of transformations first to make them much simpler.

Subtract $(5.89,40)$ from each point and divide each $y$ value by $10$ to get:

$$(0,0)\\ (1.06,1)\\ (2.33,2)\\ (3.42,3) $$

Substitute these points into $y=ax^3+bx^2+cx+d$ to get your simpler set of simultaneous equations (with $d=0$ immediately!).

Once you've solved to get $a, b$ and $c$, undo the transformations: multiply all of $a,b,c$ by $10$, add $40$ to the equation and swap $x$ for $(x-5.89)$ .

(I hope that transformations of graphs are included in "Maths B calculus"!)

Edit: help with the simultaneous equations

Plugging the values into $y=ax^3+bx^2+cx$ we get: $$ \left\{ \begin{array}{c} 1.06^3a+1.06^2b+1.06c=1 \qquad \qquad \qquad (1\mathrm{a})\\ 2.33^3a+2.33^2b+2.33c=2 \qquad \qquad \qquad (1\mathrm{b})\\ 3.42^3a+3.42^2b+3.42c=3 \qquad \qquad \qquad (1\mathrm{c}) \end{array} \right. $$

Then I’d divide each equation by its coefficient of $c$: $$ \left\{ \begin{array}{c} 1.06^2a+1.06b+c=\frac{100}{106} \qquad \qquad \qquad (2\mathrm{a})\\ 2.33^2a+2.33b+c=\frac{200}{233} \qquad \qquad \qquad (2\mathrm{b})\\ 3.42^2a+3.42b+c=\frac{300}{342} \qquad \qquad \qquad (2\mathrm{c}) \end{array} \right. $$ Now we can eliminate $c$ by doing equation $(2\mathrm{c})$ minus $(2\mathrm{b})$ and $(2\mathrm{c})$ minus $(2\mathrm{a})$, to get: $$ \begin{align} \left\{ \begin{array}{c} 1.1236a+2.36b&=-0.0662 \qquad \qquad \qquad &(3\mathrm{a})\\ 5.4289a+1.09b&=0.0188 \qquad \qquad \qquad &(3\mathrm{b}) \end{array} \right. \end{align} $$ ... where I’ve rounded the fractions on the right to $4$d.p.

I’ll leave this system of two equations for you to solve now. Once you’ve got $a$ and $b$, plug them into $(1\mathrm{a})$ to find $c$. Good luck!