Sketching a function from its sign and limits

Problem

Sketch a possible graph of a function $f$ with these features:

• negative for $x<-8$, positive for $-8<x<1$, negative for $x>1$;
• $\displaystyle\lim_{x\to-8^-}f=-\infty$, $\displaystyle\lim_{x\to-8^+}f=4$, $\displaystyle\lim_{x\to1^-}f=9$, $\displaystyle\lim_{x\to1^+}f=-9$, $\displaystyle\lim_{x\to-\infty}f=-4$, $\displaystyle\lim_{x\to+\infty}f=-\infty$.

Solution

The graph (not unique) can be built as follows:

• as $x\to-\infty$ the curve tends to the horizontal asymptote $y=-4$ while staying negative; approaching $x=-8$ from the left it drops to $-\infty$: $x=-8$ is a vertical asymptote (from the left).
• to the right of $-8$ the function restarts from value $4$ (finite right limit, jump discontinuity at $x=-8$) and stays positive, rising to $9$ as $x\to1^-$.
• at $x=1$ there is a second jump: from $9$ to $-9$; for $x>1$ the function is negative and tends to $-\infty$ as $x\to+\infty$. The two jump discontinuities are at $x=-8$ (with a left vertical asymptote) and at $x=1$; the only horizontal asymptote is $y=-4$ at $-\infty$. $\boxed{\text{Vertical asymptote }x=-8\ (\text{from the left}),\ \text{horizontal }y=-4\ (x\to-\infty),\ \text{jumps at }x=-8,\,1}$