Question

Describe the vertical asymptote(s) and hole(s) for the graph of . y= (x-3) (x-1)/(x-1)(-5)

Answer 1

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.

Step-by-step explanation:

The function is:

`y=((x-3)(x-1))/((x-1)(x-5))`

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

`\lim_(x \to \ 5+) ((x-3)(x-1))/((x-1)(x-5))=((x-3))/((x-5)) = +\infty\\ \\ \lim_(x \to \ 5-) ((x-3)(x-1))/((x-1)(x-5))=((x-3))/((x-5)) = -\infty`

Hence, the vertical line x = 5 is an asymptote.

2) Limit when x → 1

`\lim_(x \to \ 1) ((x-3)(x-1))/((x-1)(x-5))=((x-3))/((x-5)) = 1/2`

Hence, there is a hole at x = 1.

See the graph attached.