Combine an algebraic range with an intersecting line

The Teacher completes the Student blanks with the full chain

$\begin{aligned}y&=\frac{x^2+3x-1}{x-2},\\xy-2y&=x^2+3x-1,\\x^2+(3-y)x+(2y-1)&=0.\end{aligned}$

For a real input, the discriminant is nonnegative:

$\begin{aligned}(3-y)^2-4(1)(2y-1)&\ge0,\\y^2+9-6y-8y+4&\ge0,\\y^2-14y+13=(y-1)(y-13)&\ge0.\end{aligned}$

Thus $y\le1$ or $y\ge13$. Independently, substituting the forbidden input 2 into the recovered quadratic gives 9 for every $y$, so a recovered root can never be 2. The discriminant condition is sufficient too; both endpoints are attained at the turning points below.

Both copies give

$y=x+5+\frac9{x-2},$

and label asymptotes $x=2$, $y=x+5$, turns $(-1,1),(5,13)$, $y$-intercept $1/2$, and rounded $x$-intercepts −3.30 and 0.303. Independently solving $x^2+3x-1=0$ gives exact roots $(-3\pm\sqrt{13})/2$; substituting 0 gives $y=1/2$. Derivative $1-9/(x-2)^2$ changes $+\to-$ at −1 and $-\to+$ at 5, verifying the local maximum/minimum and the range endpoints. The numerator at 2 is 9, so the pole is genuine.

The source line is $l:y=m(x-2)+7$, so every member passes through $A=(2,7)$ with gradient $m$. This point is the intersection of $C$’s two asymptotes, not a point on $C$. To meet both branches in two distinct points, the source says the line must be steeper than $y=x+5$, whose gradient is 1. Its conclusion is $m>1$; this also ensures a positive gradient. The added blue line in the reconstruction is the illustrative choice $m=2$, not another printed numerical condition.

Independently put $t=x-2\ne0$. Then $C$ is $y=t+7+9/t$ and the line is $y=mt+7$. Equating gives

$\frac9t=(m-1)t,\qquad(m-1)t^2=9.$

If $m>1$, $t=\pm3/\sqrt{m-1}$ gives two distinct nonzero real inputs, one on each side of the pole. If $m=1$ the equation is impossible; if $m<1$ its left side is nonpositive. Thus the source condition is necessary and sufficient.