ok, for the vertical asymptote to the either ±2, that simply means, the denominator =0 must give thus roots, thus [tex]\bf x^2-4=0\implies x^2=4\implies x=\pm \sqrt{4}\implies x=\pm2[/tex]
so, that'd be the denominator
for the horizontal asymptote to be 4, or 4/1, we can simply make both numerator and denominator the same degree, and with leading coefficient of 4/1 respectively thus [tex]\bf \cfrac{4x^2}{1x^2-4}\impliedby HA=\cfrac{4}{1}[/tex]
now, that equation must have a point of ( 0 , -3 ), so the graph has to have such point, meaning, when y = -3, x = 0, so let's do that then, and let us add a coefficient of "a"
[tex]\bf y=\cfrac{4x^2+a}{x^2-4}\quad
\begin{cases}
y=-3\\
x=0
\end{cases}\implies \cfrac{-3=4(0)^2+a}{0^2-4}\implies -3=\cfrac{a}{-4}
\\\\\\
12=a\qquad thus\implies \boxed{y=\cfrac{4x^2+12}{x^2-4}}[/tex]
