Finding values of k for different points of intersection

You have done most of the work. As you have computed, if $x$ is an interesection point then we must have
$$x^2-(2+k)x+4=0.$$

Since we do not want ot have any intersection point, we do not want the above quadratic equation to have any solutions. Hence we must have $\Delta =b^2-4ac <0$. Thus 

\[\Delta = (-(2+k))^2 -4\times 1 \times 4=(2+k)^2-16=(k+6)(k-2)<0.\]
So we want 
\[(k+6)(k-2)<0.\]
The expression $(k+6)(k-2)$ is a continuous function and becomes zero only at $k=-6$ and $k=2$. So on the intervals 

\[(-\infty, -6), (-6, 2), \text{and}  (2, \infty)\]
it will be always positive or always negative. To determine the signes we can pick any point inside these intervals and figure out the signes 

- For $k=-10 \in (-\infty, -6)$ we have $(k+6)(k-2)=+48>0$. Hence $(k+6)(k-2)>0$ on $(-\infty, -6)$. 

- For $k=0 \in (-6, 2)$ we have $(k+6)(k-2)=-12<0$. Hence $(k+6)(k-2)<0$ on $(-6, 2)$. 

- For $k=3 \in (2, \infty)$ we have $(k+6)(k-2)=+9>0$. Hence $(k+6)(k-2)>0$ on $(2, \infty)$. 


Thus $(k+6)(k-2)<0$ only on the interval $(-6, 2)$.

In general, $(x-a)(x-b)$ is negative for $x$ between the two roots $a$ and $b$.