Question 1187367
Factor of 1 occurs, so f has a missing point instead.


{{{(5x-15)/(x-3)}}}

{{{5((x-3)/(x-3))}}}

{{{5*1}}}

{{{5}}}-------------BUT x cannot be 3 because f(x) cannot accept that input value.


f will be 5 everywhere except is empty (no point there) for x=3.