We are given
[tex] y=\frac{x^2-4x+3}{x^2-9} [/tex]
Vertical asymptotes:
Firstly, we will factor numerator and denominator
we get
[tex] y=\frac{(x-1)(x-3)}{(x-3)(x+3)} [/tex]
We can see that (x-3) is common in both numerator and denominator
so, we will only set x+3 to 0
and then we can find vertical asymptote
[tex] x+3=0 [/tex]
[tex] x=-3 [/tex]
Hole:
We can see that (x-3) is common in both numerator and denominator
so, hole will be at x-3=0
[tex] x-3=0 [/tex]
[tex] x=3 [/tex]
Horizontal asymptote:
We can see that degree of numerator is 2
degree of denominator is also 2
for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator
and we get
[tex] y=\frac{1}{1} [/tex]
[tex] y=1 [/tex]
so, option-D............Answer
