First of all, the linear interpolation function is usually called "lerp", not "learp", so that's what I'll call it.
You have the patch equation, which you wrote in terms of the "lerp" function. Now just use the definition of lerp:
$$
\text{lerp}(t, \mathbf{P}, \mathbf{Q}) = (1-t)\mathbf{P} + t\mathbf{Q}
$$
This allows you to write out your patch equation in a more algebraic form; it becomes:
$$
\mathbf{X}(u,v) =
(1-v)\left[(1-u)\mathbf{P}_0 + u\mathbf{P}_1\right] +
v\left[(1-u)\mathbf{P}_2 + u\mathbf{P}_3\right]
$$
Multiplying out gives
$$
\mathbf{X}(u,v) = (1-u)(1-v)\mathbf{P}_0 + u(1-v)\mathbf{P}_1
+ (1-u)v\mathbf{P}_2 + uv\mathbf{P}_3
$$
Now just take partial derivatives with respect to $u$ and $v$ to get tangent vectors, and form the cross product to get the surface normal.
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
Hint: You are given four pairs $(x_i,y_i)$ and you are told $f(x_i)=y_i$ for each pair. So just suppose that $$f(x)=ax^3+bx^2+cx+d$$ Then substitute each pair of numbers into this equation to get four equations in the four unknowns $a$, $b$, $c$, and $d$. Solve the system of equations for these four unknowns and you should get $a=-0.24728$, $b=0.57093$, $c=0.17636$, and $d=2.7$, approximately.
For example, when you substitute the first pair, you get $$2.7=d$$ which is especially nice. I would immediately replace $d$ with $2.7$ in the remaining three equations to get a system of three equations in three unknowns $a$, $b$, and $c$ to solve (now you already have $d$). When you substitute the second pair, and the known value of $ d $, you get $$2.9=(0.5)^3a +(0.5)^2b+0.5c+2.7$$ which can be rewritten as $$0.125a+0.25b+0.5c=0.2$$ and so on.
I suppose you could use this equation to write one of the variable in terms of the remaining two, such as $$c=0.4-0.25a-0.5b.$$
Can you take it from here? Do you need help solving the system of equations?