Question

1. The vertical asymptote is x=12. There is a "hole" in the graph at x=13. The graph has the oblique asymptote y=x-14. The horizontal asymptote is y=1

Answer 1

Given,

The rational function is,

$f(x)=\frac{x^2-1}{x-1}$

As the degree of numerator is greater than the degree of denomenator, then vertical asymptote is not possible.

For hole in the graph,

Here, in the function,

$f(x)=\frac{(x^{}-1)(x+1)}{x-1}$

The value (x-1) is cancel out from numerator and denomenator

So, the graph is discontinious at x=1

Hence, there is a "hole" in the graph at x=1.

Now, for the oblique asymptote,

Dividing the numerator by denominator,

On dividing we get (x + 1),

Hence, the function have oblique asymtote at y=x+1