Find exact input and output ranges

The source requires $x-3>0$, hence $x>3$. Its graph labels vertical asymptote $x=3$ and intercept $(4,0)$, and gives $y\in\mathbb R$. Independently, for every real $y$, the input $x=3+e^y>3$ gives that output; this proves the full range rather than just suggesting it from a window.

The printed curve is $y=-\sqrt{3-x}$, with the radical over the whole $3-x$. Require $3-x\ge0$, so $x\le3$. The source graph labels $(3,0)$ and $(0,-\sqrt3)$; its range is $y\le0$. Independently, for every $y\le0$, choose $x=3-y^2$ to recover $y=-\sqrt{y^2}=-|y|=y$. The endpoint is included.

For $y=(x^2+2)/(x+1)$, the source gives $x\ne-1$, or $x<-1$ or $x>-1$. At the excluded input the numerator is $3$, so this is not a removable cancellation.

The Teacher completes the Student working space. Starting from

$y=\frac{x^2+2}{x+1},\qquad xy+y=x^2+2,$

write $x^2+(-y)x+(2-y)=0$. The printed quadratic-formula chain is

$\begin{aligned}&x\\ & =\frac{-(-y)\pm\sqrt{(-y)^2-4(1)(2-y)}}{2(1)}\\ & =\frac{y\pm\sqrt{y^2+4y-8}}2.\end{aligned}$

For real $x$, its discriminant must be nonnegative. This gives an exact algebraic condition on the possible outputs.

The Teacher gives $y^2+4y-8\ge0$ and solves the boundary equation:

$y^2+4y-8=0,\qquad (y+2)^2-12=0,\qquad y=-2\pm\sqrt{12}.$

Its upward parabola shows that the required values lie outside the roots, including the roots: $y\le-2-\sqrt{12}$ or $y\ge-2+\sqrt{12}$. Independently, the recovered quadratic has value $3$ at $x=-1$ for every $y$, so no recovered root is the excluded input. Thus the discriminant condition is sufficient as well as necessary, and both endpoints are attained.

The shared graphical method is labelled for non-exact answers. It labels $x=-1$, $y=x-1$, $(0,2)$, maximum $(-2.73,-5.46)$ and minimum $(0.732,1.46)$, then reports $y\le-5.46$ or $y\ge1.46$. Retain these as rounded readings, not exact boundary inequalities.

Independently, division gives $y=x-1+3/(x+1)$ and $y'=1-3/(x+1)^2$. The exact turning inputs are $-1\pm\sqrt3$, with ordinates $-2\pm2\sqrt3=-2\pm\sqrt{12}$. They agree with the source’s rounded labels. The source recommends using known input/output restrictions to choose Xmin, Xmax, Ymin and Ymax so relevant graph features can be seen; a finite display cannot literally show an unbounded whole curve.