Given four points $(x_i,y_i)$ consider the functions $$f_1(x)=\frac {(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$ so that $f_1(x_1)=1$ and $f_1(x_i)=0, i\neq 1$, and similarly $f_2, f_3, f_4$. Note that the $f_i$ are cubic in $x$.
Then $p(x)=y_1f_1(x)+y_2f_2(x)+y_3f_3(x)+y_4f_4(x)$ is at most a cubic polynomial and passes through the four given points.
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!
Best Answer
First of all, since the dilation factor is equal to $1$, the equation automatically reduces to $y=(x-b)^3+c$. Now, since the equation only has a single factor, $(x-b)$, the vector $(b,c)$ is the result we're looking for. Now, the $b$ and $c$ can be found by plugging in the given intercepts: $$0=(-4-b)^3+c$$$$28=(-b)^3+c$$ Since this is a system of nonlinear equations, we can solve it by substitution. $$c=-(-4-b)^3=(4+b)^3$$$$28=(-b)^3+(4+b)^3$$ Since we can't easily combine the right-hand side of this in factored form, we'll expand it. $$28=64+48b+12b^2+b^3-b^3=64+48b+12b^2$$ Well now, that's just a simple quadratic. Solving it, we get two solutions: $(-1, 0)$ and $(-3, 0)$. Let's try plugging both of those back into the first equation: $$c=(4-1)^3=27$$$$c=(4-3)^3=1$$ So now we have two possible coordinate pairs, $(-1, 27)$ and $(-3, 1)$. If you graph these equations, you'll see that either coordinate works.