[tex]\bf g(x)=\cfrac{x^2}{x^2-2x-3}[/tex]
so, notice, the degree of the numerator is the same degree as the denominator's, therefore the horizontal asymptote is at
[tex]\bf g(x)=\cfrac{1x^2}{1x^2-2x-3}\implies \cfrac{1}{1}\implies \stackrel{\textit{horizontal asymptote }y=1}{1}[/tex]
the vertical asymptotes are at the zeros of the denominator, let's check.
[tex]\bf x^2-2x-3=0\implies (x-3)(x+1)=0\implies x=
\begin{cases}
3\\
-1
\end{cases}[/tex]
there are no oblique or slanted asymptotes, that only occurs when the numerator's degree is higher by 1.
to get the x-intercept, simple enough, just set y = 0 and solve for "x" as usual.
[tex]\bf \stackrel{0}{g(x)}=\cfrac{x^2}{x^2-2x-3}\implies 0=\cfrac{x^2}{x^2-2x-3}\implies \boxed{0=x}[/tex]
and to get the y-intercept, set x = 0, and solve for "y".
[tex]\bf g(x)=\cfrac{0^2}{0^2-2(0)-3}\implies g(x)=\cfrac{0}{-3}\implies \boxed{g(x)=0}[/tex]
