Connect a derivative discriminant to the missing range

$y=x+a+(a^2+a)/(x-a)$. Its oblique asymptote is y=x+a; comparison with y=x+2 gives a=2.

$y'=[2x(x-a)-(x^2+a)]/(x-a)^2=(x^2-2ax-a)/(x-a)^2$. Setting this equal to zero gives $x^2-2ax-a=0$. Two turns require $(-2a)^2-4(1)(-a)>0$, hence $4a^2+4a>0$, or $a(a+1)>0$. Therefore $\{a\in\mathbb R:a<-1\text{ or }a>0\}$. In these intervals a²+a>0, so neither derivative zero is the excluded input x=a; the simple zeros change derivative sign.

For a=2, $y=(x^2+2)/(x-2)$, x≠2. Then $y(x-2)=x^2+2$, giving $x^2-yx+2y+2=0$. To have no real x, require $(-y)^2-4(1)(2y+2)<0$, or $y^2-8y-8<0$.

The boundary roots are $[8\pm\sqrt{(-8)^2-4(1)(-8)}]/[2(1)]=(8\pm\sqrt{96})/2=(8\pm4\sqrt6)/2=4\pm2\sqrt6$. Thus $4-2\sqrt6<y<4+2\sqrt6$ is missing, with $y_1=4-2\sqrt6$ and $y_2=4+2\sqrt6$.

$y=x+2+6/(x-2)$, x≠2. The asymptotes are x=2 and y=x+2. The stationary points are $(2-\sqrt6,4-2\sqrt6)$, a maximum, and $(2+\sqrt6,4+2\sqrt6)$, a minimum. The y-intercept is (0,−1); x²+2>0 rules out x-intercepts.