Use missing output levels and a circle comparison to count roots

The vertical asymptote gives b=1. The quotient rule yields $y'=[(2x+a)(x+1)-(x^2+ax+4)]/(x+1)^2=(x^2+2x+a-4)/(x+1)^2$. At the stationary point x=2, $(4+4+a-4)/(2+1)^2=0$, hence a=−4.

$y=(x^2-4x+4)/(x+1)=x-5+9/(x+1)$. Therefore the other asymptote is y=x−5.

$y(x+1)=x^2-4x+4$ gives $yx+y=x^2-4x+4$, hence $x^2+(-4-y)x+4-y=0$. For no real x, $(-4-y)^2-4(1)(4-y)<0$. Thus $y^2+8y+16-16+4y<0$, so $y^2+12y<0$, or y(y+12)<0. The unattainable set is exactly −12<y<0. The excluded x=−1 never solves this quadratic, since substitution gives 9.

Asymptotes are x=−1 and y=x−5. Intercepts are (0,4) and (2,0), with the latter a local minimum. The other stationary point is a local maximum (−4,−12), as shown in the source. In fact $y'=(x+4)(x-2)/(x+1)^2$. Boundary outputs −12 and 0 are attained, explaining why they are not in the missing-output interval.

For x≠−1, divide the equation by (x+1)²to obtain $[(x^2-4x+4)/(x+1)]^2=4-x^2$. With y equal to the curve, this is $y^2=4-x^2$, or $y=\pm\sqrt{4-x^2}$: the circle x²+y²=4. The source circle sketch labels (−2,0),(2,0),(0,2),(0,−2) and shows two intersections with C. Hence the original equation has two distinct real roots. x=−1 is not a lost root: the original sides there are 0 and 81.

As a supplementary check, bring all terms to one side and factor to get $(x-2)(2x^3-2x^2+17x-6)=0$, up to an overall nonzero sign. The cubic derivative $6x^2-4x+17$ is positive for all real x, so it has exactly one real root; its values at 0 and 1 have opposite signs. That root is distinct from x=2. This confirms the source’s two-intersection deduction without replacing it.