Solution
We are given the function
[tex]f(x)=\frac{x^2-4}{3x^2+x-4}[/tex]First, to get the Vertical asymptotes, we only need to equate the denominator to zero and then solve for x
[tex]\begin{gathered} 3x^2+x-4=0 \\ \\ 3x^2-3x+4x-4=0 \\ \\ 3x(x-1)+4(x-1)=0 \\ \\ (x-1)(3x+4)=0 \\ \\ x=1,-\frac{4}{3} \end{gathered}[/tex]Therefore, the Vertical Asymptotes are
[tex]\begin{gathered} x=1 \\ and \\ x=-\frac{4}{3} \end{gathered}[/tex]To get the horizontal asymptotes, we only need to take the highest power of x from both numerator and denominator
[tex]\begin{gathered} y=\frac{x^2}{3x^2} \\ \\ y=\frac{1}{3} \end{gathered}[/tex]Therefore, the horizontal asymptotes is
[tex]y=\frac{1}{3}[/tex]