Answer:
Hole=1 and vertical asymptote=5
Step-by-step explanation:
It is given that:
[tex]Y=\frac{(x-3)(x-1)}{(x-1)(x-5)}[/tex]
Look at the denominator and determine the values for which it is zero, therefore
(x-1)(x-5)=0⇒x=1,5
There is a common factor of (x-1), so we can cancel that common factor:
[tex]Y=\frac{x-3}{x-5}[/tex]
But note we can only do that provided x-1≠0, x≠1.
We have already got that when x=1 the denominator is zero, so there is a removable discontinuity at that point (or a "hole").
when x=5, he denominator is also zero and this is the only vertical asymptote.
