Given the rational function;
[tex]y=\frac{x^2+7}{2x^2+x-3}[/tex]
The vertical asymptotes are the points where the denominator is zero;
Thus, we have;
[tex]2x^2+x-3=0[/tex]
Then,
[tex]\begin{gathered} 2x^2-2x+3x-3=0 \\ 2x(x-1)+3(x-1)=0 \\ 2x+3=0\text{ or x-1 = 0} \\ x=-\frac{3}{2}\text{ or x = 1} \\ \end{gathered}[/tex]
CORRECT OPTION: D
