For this problem, we need to write the equation to a function that has a discontinuity at x = 1.
First, we need to find one that has a hole at x = 1. This is shown below:
[tex]y=\frac{(x+2)(x-1)}{(x-1)}[/tex]This function is not defined at the point x = 1, therefore it will have a whole on that point.
Then we need to find a jump discontinuity, meaning that the function will abruptly change from one value to the other on that point. We have:
[tex]y=\begin{cases}x+2,\text{ for }x<1{} \\ x^2,\text{ for }x\geqslant{1}\end{cases}[/tex]We need to determine a function with a vertical asymptote in 1. For that, we need to use a rational function with a denominator equal to "x-1".
[tex]y=\frac{1}{x-1}[/tex]