Use long division for an oblique asymptote

The source identifies $x=-1$ as undefined. Independently the numerator there is $4\ne0$; the quotient is unbounded as $x$ approaches $-1$, so $x=-1$ is a vertical asymptote.

The Teacher annotation divides $x^2-3x$ by $x+1$: quotient term $x$, subtract $x^2+x$ to leave $-4x$; next quotient term $-4$, subtract $-4x-4$ to leave remainder $4$. Therefore $y=x-4+\frac4{x+1}.$ The printed shorthand says $y\to x-4$. Precisely, $y-(x-4)=4/(x+1)\to0$ as $x\to\pm\infty$, so $y=x-4$ is the oblique asymptote.

The source labels the origin, $(3,0)$, the maximum $(-3,-9)$ and minimum $(1,-1)$. The Teacher copy adds dashed $x=-1$ and $y=x-4$. Independently, $y=x(x-3)/(x+1)$ supplies the intercepts, and $y^{\prime}=1-4/(x+1)^2$ vanishes at $-3,1$. Its sign changes $+\to-$ at $-3$ and $-\to+$ at $1$, verifying the printed turning points. The left branch is below its oblique line and tends to $-\infty$ at $-1^-$; the right lies above it and tends to $+\infty$ at $-1^+$. Example 4 next develops turning-point methods in detail; its full continuation remains separate.