Rational curve: coefficients, hyperbola comparison and stationary points

The points give $9a+3b+c=23$ and $4a+2b+c=10$. The derivative is $\dfrac{ax^2-2ax-b-c}{(x-1)^2}$; stationarity at $x=2$ gives $b+c=0$. Solving these three equations (the source uses a GC) gives $a=3,b=-2,c=2$. The lecturer guidance in the Conics Teacher copy (PDF 28, printed 140) recommends using a system of linear equations (SOLE) for these three conditions. Set up all three equations before solving for $a,b,c$; reviewing simultaneous equations in advance helps. This makes the source’s GC solution a system-solving step rather than an unexplained coefficient guess.

Long division gives $y=3x+1+\dfrac3{x-1}$. The asymptotes are $x=1$ and $y=3x+1$. Since $y^\prime=3-3/(x-1)^2$, stationary inputs are $x=0,2$, giving a maximum $(0,-2)$ and minimum $(2,10)$. The sole axial intersection is $(0,-2)$: $3x^2-2x+2=0$ has negative discriminant $4-24=-20$, so no real $x$-intercepts.

The comparison hyperbola has centre $(0,1)$ and asymptotes $y=1\pm(m/k)x$. Comparing the positive-gradient asymptote with $y=3x+1$ gives the source answer $0<m/k\le3$, including equality. For a check beyond the sketch, put $r=m/k$. At an intersection, $y-1=3x+3/(x-1)$ and $r^2=(y-1)^2/(x^2-k^2)$. If $x<0$ or $x>1$, $|y-1|>3|x|$. For $0<x<1$, $|y-1|-3x=3/(1-x)-6x>0$ because $1-2x+2x^2>0$. Thus every intersection requires $r>3$. Conversely for any fixed $k>0$, on $x>\max(k,1)$ this ratio is continuous, tends to infinity at the left boundary and to $3$ as $x\to\infty$, so every $r>3$ produces an intersection. Therefore $0<r\le3$ is exact.

With $a=b=-1$, $y^\prime=\dfrac{-x^2+2x+1-c}{(x-1)^2}$. For no stationary input, the numerator quadratic must have negative discriminant: $2^2-4(-1)(1-c)=8-4c<0$, so $c>2$. For $c<2$, its two roots $1\pm\sqrt{2-c}$ are in the domain. The source separately notes that at $c=2$, $y=\dfrac{(x-1)(-x-2)}{x-1}=-x-2$ for $x\ne1$, also with no stationary point, but that case is explicitly excluded in the question.