Sketch rational curves from division and domain

Both sources give

$y=\frac{2x-3}{x-4}=2+\frac5{x-4}.$

The asymptotes are $y=2$ and $x=4$; intercepts follow from $x=0\Rightarrow y=3/4$ and $y=0\Rightarrow x=3/2$. The Teacher adds labels to the existing Student curve. The source says there are no stationary points; independently $y'=-5/(x-4)^2<0$ on each domain branch. The numerator at $x=4$ is nonzero, so this is a pole, not a hole.

The printed division chain is

$\begin{aligned}&y\\ & =\frac{x^2-2}{x^2-1}\\ & =1-\frac1{x^2-1}\\ & =1-\frac1{(x-1)(x+1)}.\end{aligned}$

Asymptotes are $y=1$ and $x=\pm1$. Substituting $x=0$ gives $(0,2)$; solving $x^2-2=0$ gives $(\pm\sqrt2,0)$. The Teacher completes the graph labels and the missing left vertical asymptote in the visible Student sketch. “Use G.C.” is the printed stationary-point instruction. Independently

$y'=\frac{2x}{(x^2-1)^2}.$

On the central branch the derivative changes $-\to+$ at 0, so $(0,2)$ is a local minimum; neither exterior branch has a stationary point. It is not a global minimum over the disconnected domain. Both denominator zeros are genuine poles.

The Teacher completes the Student blanks with

$\begin{aligned}&y\\ & =\frac{2x^2-x+1}{x-2}\\ & =2x+3+\frac7{x-2}.\end{aligned}$

Its asymptotes are $y=2x+3$ and $x=2$. At $x=0$, $y=-1/2$. The equation $2x^2-x+1=0$ has no real solutions (independently its discriminant is $1-8=-7$), so there are no $x$-intercepts. The source asks for G.C. stationary points and labels a maximum $(0.129,-0.483)$ and minimum $(3.87,14.5)$ in the Teacher graph. Independent calculus gives

$y'=2-\frac7{(x-2)^2},\qquad x=2\pm\sqrt{7/2},\qquad y=7\pm2\sqrt{14}.$

The minus signs give the left local maximum; the plus signs give the right local minimum. The original pole at 2 separates them. Keep the exact expressions for calculation and treat the printed decimals as rounded.

The source compares ZOOM 6: ZStandard with adjusted window settings: the standard view misses the right-hand branch, while the adjusted view shows both sides of $x=2$. A finite view can show relevant features, never literally all of an unbounded graph. It also warns that the left maximum may seem to lie on the $y$-axis. Use ZOOM 1: ZBox to enlarge that region. The source maximum screen reports approximately $X=0.12917$ and $Y=-0.483315$, not $x=0$. Independently the exact left-turn input is $2-\sqrt{7/2}\approx0.1291713>0$; at $x=0$ the curve is still increasing, with derivative $1/4$.

The Teacher fills the visible Student blanks:

$\begin{aligned}&y\\ & =\frac{x^2-x-12}{x-2}\\ & =x+1-\frac{10}{x-2}.\end{aligned}$

The asymptotes are $y=x+1$ and $x=2$. At $x=0$, $y=6$; setting $x^2-x-12=(x+3)(x-4)=0$ gives $(-3,0)$ and $(4,0)$. The source says no stationary points. Independently $y'=1+10/(x-2)^2>0$ on each branch, so the curve is increasing on each side of the pole, with no turns.

Both sources give

$\begin{aligned}&y\\ & =\frac{(x+5)^2}{x^2-5x}\\ & =1+\frac{15x+25}{x(x-5)}.\end{aligned}$

Thus $y=1$ is the horizontal asymptote and $x=0,5$ are vertical asymptotes. At $x=0$ the function is undefined, so there is no $y$-intercept. The only $x$-intercept is $(-5,0)$, where the squared numerator gives contact. The printed instruction is “Use G.C.”; the Teacher graph also labels $(5/3,-8)$. Independently

$\begin{aligned}&y'\\ & =-\frac{5(x+5)(3x-5)}{x^2(x-5)^2}.\end{aligned}$

Its sign gives a local minimum $(-5,0)$ on the left branch and a local maximum $(5/3,-8)$ on the central branch. On the far-right branch the derivative is negative, with no turn. Neither turn is a global extremum over all three branches.

The shared source warning says ZOOM 6: ZStandard does not clearly show the curve, and the left tail may appear to approach 0. This is false: the division form shows $y\to1$ as $x\to-\infty$, since $(15x+25)/(x(x-5))\to0$. Determine asymptotes and intercepts analytically before using a calculator for shape. The Teacher also recommends showing long division if needed; compare each numerator with denominator × quotient + remainder before plotting.