Find absent output levels from a quadratic discriminant

Write y=k. No intersection means $k=(-9x^2+6x-1)/(x-1)$ has no real root. Thus $-9x^2+6x-1=kx-k$, or $9x^2+(k-6)x+1-k=0$. The discriminant condition is $(k-6)^2-4(9)(1-k)<0$, giving $k^2+24k<0$, hence k(k+24)<0 and −24<k<0. Therefore the missing output interval is −24<y<0.

Subtract $-9x(x-1)=-9x^2+9x$ from the numerator to leave −3 x−1. Subtract $-3(x-1)=-3x+3$ to leave −4. Thus $y=-9x-3-4/(x-1)$, with asymptotes x=1 and y=−9 x−3.

The y-intercept is (0,1). The numerator is−(3 x−1)², so the x-intercept is (1/3,0), a local minimum; the source labels it(0.333,0). The other turn is a local maximum (5/3,−24), source rounding(1.67,−24). These follow from $y'=-9+4/(x-1)^2$. Both boundary output values are attained.