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The conditions are:

• $\lim_{x\rightarrow\infty}f(x)=1$
• $\lim_{x\rightarrow-\infty}f(x)=2$
• $\lim_{x\rightarrow-2^-}f(x)=\infty$
• $\lim_{x\rightarrow-2^+}f(x)=\infty$
• $\lim_{x\rightarrow1^-}f(x)=-\infty$
• $\lim_{x\rightarrow2^+}f(x)=\infty$
• $\lim_{x\rightarrow3^-}f(x)=2$
• $\lim_{x\rightarrow3^+}f(x)=4$

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1 Answer

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Well, given

• $\lim_{x\rightarrow\infty}f(x)=1$
• $\lim_{x\rightarrow-\infty}f(x)=2$
• $\lim_{x\rightarrow-2^-}f(x)=\infty$
• $\lim_{x\rightarrow-2^+}f(x)=\infty$
• $\lim_{x\rightarrow1^-}f(x)=-\infty$
• $\lim_{x\rightarrow2^+}f(x)=\infty$
• $\lim_{x\rightarrow3^-}f(x)=2$
• $\lim_{x\rightarrow3^+}f(x)=4$

The first two tell you what the end behavior is, i.e., there is a horizontal asymptote at $y=1$ to the left and $y=2$ to the right.

The third and fourth indicate a vertical asymptote at $x=-2$ that is approaching $\infty$ from both left and right of $x=-2$.

The fifth indicates a vertical asymptote toward $-\infty$ approaching $x=1$ from the left.

The sixth indicates a vertical asymptote toward $\infty$ approaching $x=2$ from the right.

The last two indicate a jump discontinuity where $f(x)$ approaches $2$ from the left of $x=3$ and $4$ from the right of $x=3$.

If you start sketching approximations of all of these together you should get a clear picture of your function.

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