In the denominator of the fraction is given expression [tex](x-1)(x-5).[/tex] This means that
[tex]x\neq 1,\\x\neq 5.[/tex]
Now simplify the expression [tex]y=\dfrac{(x-3)(x-1)}{(x-1)(x-5)}:[/tex]
[tex]y=\dfrac{(x-3)(x-1)}{(x-1)(x-5)}=\dfrac{x-3}{x-5}.[/tex]
This means that [tex]x=5[/tex] is the equation of vertical asymptote and, therefore, since function id not defined for [tex]x=1[/tex] this point is hole of the function.
