Justify vertical and horizontal asymptotes

The source identifies $x=3$ as undefined. For $x\to3^+$, $x-3$ is small and positive, so $1/(x-3)\to+\infty$ and $y\to+\infty$. For $x\to3^-$ it is small and negative, giving $y\to-\infty$. Thus $x=3$ is a vertical asymptote. These one-sided limits justify it; an undefined input alone could instead be a removable hole.

As $x\to\pm\infty$, $1/(x-3)\to0$, so $y\to1$. Hence the horizontal asymptote is $y=1$. The Teacher note warns that a calculator has limited capabilities for finding asymptotes: establish the limits analytically.

The printed sketch labels $(2,0)$ and $(0,2/3)$. Independently, solve $1+1/(x-3)=0$ to get $x=2$ and substitute $x=0$ to get $2/3$. Draw the asymptotes first using dotted lines, as the next-page source remark directs. The Teacher copy adds the dashed lines and labels $x=3,y=1$. The right branch lies above $y=1$ and the left branch below it; neither crosses these asymptotes for this function. The QR invitation to explore asymptotes is retained in the original artwork; no remote activity is asserted reviewed.