Answer:
The function f(x) = x^2-6x+5/x^2-3 is discontinuous at [tex]x= \sqrt{3}[/tex] and [tex]x= -\sqrt{3}[/tex]
Step-by-step explanation:
The function is discontinuous at the roots of denominator [tex]x^{2} -3=0[/tex] which are [tex]x= \sqrt{3}[/tex] and [tex]x= -\sqrt{3}[/tex].
The function f(x) = x^2-6x+5/x^2-3 has also two vertical asymptotes at these points.
The Graph of the function is attached. One horizontal asymptote can also be identified at y=1
