Intuition behind 2nd order approximation, help please

I know how to apply the formula for Taylor Expansions. But what I want to understand is the intuition. Let me explain with the following example:

If $y=x^5$ its 1st derivative is $5x^4$ and its 2nd derivative is $20x^3$.

If we do a 1st order approximation at (2) we get that the equation $y=80x-128$ does the job. Instead of directly using the Taylor formula I evaluated y and it's two derivatives at (2) and got:

\begin{align*} Y&=(2)^5 =32 \\ Y&=5 \cdot (2)^4 = 80 \\ Y&=20 \cdot (2)^3 =160 \end{align*}

As you can see the 1st derivative gives us a slope of 80 at (2). So I set $32=80x-b$ in order to get the intercept and thus a linear equation that approximates $x^5$ at (2).

If we want a 2nd order approximation we can use the Taylor formula and get $$32 + 80 \cdot (x-2) + 80 \cdot (x-2)^2$$

So my question is: how can I get the 2nd order approximation by the same method I got the 1st one? I want to get a second order approximation without directly using the Taylor formula, so that I can have a better grasp at the intuition and derivation of the formula.

I thought about doing the same step with $5x^4$ at (2) and then approximating it linearly by getting the slope, by it's 1st derivative, $20x^3$ , at (2) and then trying to work up but I can't get the same result.

Any suggestions? Thanks!

The intuition that I'm aware of behind Taylor's formula is that the $n$th degree Taylor approximation of a function $f(x)$ at a point $a$ gives a polynomial function $T_n(x)$ such that $$f^{(i)}(a)=T_n^{(i)}(a),\quad\text{for }i=0,1,2,\ldots,n$$ What this says is that the Taylor polynomial shares its value and the values of its first $n$ derivatives with the function at the point $a$.
In the case of first degree approximation, your process leads you to match the value of the linear polynomial and the slope at the point $x=2$.
To get a second order approximation in a similar manner, you could begin with an arbitrary quadratic polynomial $T_2(x)=ax^2+bx+c$, and use the three conditions that its value and the value of its first two derivatives at $2$ be those of $x^5$ at $2$. This means we want to solve the system of equations $$4a+2b+c=32$$ $$4a+b=80$$ $$2a=160$$
This leads to the solution $a=80,$ $b=-240$, and $c=192$, which recovers your result.