Understanding the solution to this polynomial estimation problem Show that there exist $K, N > 0$  such that for $x ∈\mathbb{R}$,
$$x ≥ N \implies
\frac{3x^
2 − 4x + 8}{
5x + 6}
≥ Kx.$$
Solution:
For $x ≥ 4,$ we have $4x ≤ x^
2$
so that
$$3x^
2 − 4x + 8 ≥ 3x^
2 − x^
2 = 2x^
2.$$
Similarly, for $x ≥ 1,$ we have
$$5x + 6 ≤ 11x.$$
Therefore, when $x ≥ 4,$ we have
$$\frac{3x^
2 − 4x + 8}{
5x + 6}
≥\frac{
2x^
2}{
11x}
=
\frac{2}
{11} x,$$
and so we can take $N = 4$ and 
$K = \frac{2}{
11} .$
I'm struggling to follow the logical steps in this solution. I'm not sure why the inequality of $x ≥ 4$ and $x ≥ 1$ has been chosen. Can someone please break down this solution and show where the steps come from?

Essentially the method of attack that was used in this proof is to find a quadratic function that is less than the numerator and a linear function that is greater than the denominator. Then when you divide these you get a linear function which is less than the original $\frac{3x^2-4x+8}{5x+6}$. 
In layman's terms the question is asking if it is possible to find some real number $N$ such that $\frac{3x^2-4x+8}{5x+6}$ is greater than some linear function $Kx$ for all $x$ greater than or equal to $N$. 
So again we are looking for a quadratic equation which is less than $3x^2-4x+8$ for some values of $x > 0$. Notice that $3x^2 + 8 \geq 3x^2 $ for all $x > 0$. Now we need to account for the $-4x$ term still in the numerator. There's actually an infinite amount of choices we could do for the next step. In  the proof you presented the author realized that $x^2$ would be greater than or equal to $4x$ eventually (namely at $x = 4$). So from the inequality above we note that $3x^2 - 4x + 8 \geq 3x^2 - x^2 = 2x^2$ when $x \geq 4$ (because we subtracted something greater from the RHS than the LHS). 
Now for the next step again we have an infinite amount of choices, however the author realized that $5x +6 \leq 11x$ when $x \geq 1$. Of course they could just have easily said $5x+6 \leq 12x$ or an infinite number of other choices for that matter (thus finding a different value of $K$) but they chose this. Now since $5x + 6 \leq 11x$ when $x \geq 1, $ it is certainly also true when $x \geq4.$ 
Now for the final step, we can obtain the equation $\frac{3x^2 - 4x + 8}{5x+6} \geq \frac{2x^2}{11x} = \frac{2}{11}x$ for $x \geq 4$ because the numerator of the LHS is greater than the numerator of the RHS and the denominator of the LHS is less than the numerator of the RHS for $x\geq4$ as we showed above. 
Thus we can pick $N = 4$ and $K = \frac{2}{11}$. To make sure you understand it try and find another value of $N$ and $K$, for instance by realizing that $3x^2 - 4x + 8 \geq 3x^2 -2x^2 = x^2$ for $x \geq 4$ in the first step.  
