[tex]\begin{gathered} g(x)=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-3\right)} \\ \text{ In the numerator we can determine the function zeros} \\ (x-2)(x+2)=0,\text{ set the numerator equal to zero and we have} \\ x-2=0\rightarrow x=2 \\ x+2=0\rightarrow x=-2 \\ \text{ We have 2, and -2 for function zeros} \\ \text{For vertical asymptotes we set the denominator to zero} \\ (x+1)(x-3)=0 \\ x+1=0\rightarrow x=-1 \\ x-3=0\rightarrow x=3 \\ \text{ We have -1 and 3 for vertical asymptotes} \end{gathered}[/tex]
