Solution:
From the graph, we can see that there is a hole at x=-2
[tex]\begin{gathered} x=-2 \\ (x+2) \end{gathered}[/tex]
From the image below we have the vertical asymptotes to be
[tex]\begin{gathered} x=3 \\ (x-3) \end{gathered}[/tex]
The x-intercepts or the zeros is given at
[tex]\begin{gathered} x=2 \\ (x-2) \end{gathered}[/tex]
Hence,
The equation of the graph will be calculated below as
[tex]g(x)=\frac{a(x-2)(x+2)}{(x-3)(x+2)_}[/tex]
To find a, we will use the coordinates below
[tex](x,y)=(1,1)[/tex][tex]\begin{gathered} g(x)=\frac{a(x-2)(x+2)}{(x-3)(x+2)} \\ 1=\frac{a(1-2)(1+2)}{(1-3)(1+2)} \\ 1=-\frac{3a}{-6} \\ 3a=6 \\ \frac{3a}{3}=\frac{6}{3} \\ a=2 \end{gathered}[/tex]
Hence,
The equation of g(x) will be
[tex]\begin{gathered} g(x)=\frac{2(x-2)(x+2)}{(x-3)(x+2)} \\ g(x)=\frac{2(x^2-4)}{x^2-x-6} \\ g(x)=\frac{2x^2-8}{x^2-x-6} \end{gathered}[/tex]
The final answer is
[tex]\Rightarrow g(x)=\frac{2x^{2}-8}{x^{2}-x-6}[/tex]
The SECOND OPTION is the right answer
