Answer:
The vertical line x = 5 is a vertical asymptote
There is a hole at (1, 1/2)
Aslo, see the explanation below and the graph attached.
Explanation:
The function is:
[tex]y=\frac{(x-3)(x-1)}{(x-1)(x-5)}[/tex]
The vertical asymptotes happen at the points where the denominator equals zero (the function is undefined) and the limit of the function trends to + or - infinity.
If the denominator equals zero, but the limit of the function is defined, then that does not define an asymptote but a hole.
There are two points where the denominator is zero. Those are:
- x - 5 = 0 ⇒ x = 5, and
- x - 1 = 0 ⇒ x = 1
Now, calculate both limits:
1) Limit when x → 5
[tex]\lim_{x \to \ 5+} \frac{(x-3)(x-1)}{(x-1)(x-5)}=\frac{(x-3)}{(x-5)} = +\infty\\ \\ \lim_{x \to \ 5-} \frac{(x-3)(x-1)}{(x-1)(x-5)}=\frac{(x-3)}{(x-5)} = -\infty[/tex]
Hence, the vertical line x = 5 is an asymptote.
2) Limit when x → 1
[tex]\lim_{x \to \ 1} \frac{(x-3)(x-1)}{(x-1)(x-5)}=\frac{(x-3)}{(x-5)} = 1/2[/tex]
Hence, there is a hole at x = 1.
See the graph attached.
