How do you graph f(x)=x^2/x−1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Alright, So for this question we are looking for six items - holes, vertical asymptotes, horizontal asymptotes, x intercepts, and y intercepts - in the equation ${\displaystyle f\left(x\right)=\frac{x^2}{x-1}}$ First lets graph it
graph{x^2/(x-1 [-10, 10, -5, 5]}
Right off the bat you can see some strange things happening to this graph. Lets really break it down.
To begin, lets find the x and y intercept. you can find the x intercept by setting y=0 and vise versa x=0 to find the y intercept.
For the x intercept:
${\displaystyle 0=\frac{x^2}{x-1}}$
0=x
Therefore, x=0 when y=0. So without even knowing that information, we have just found BOTH the x and y intercept.
Next, lets work on the asymptotes. To find the vertical asymptotes, set the denominator equal to 0, then solve.
0=x−1
x=1
So we just found that there is a vertical asymptote at x=1. You can visually check this by looking at the above graph. Next, lets find the horizontal asymptote.
There are three general rules when talking about a horizontal asymptote.
1) If both polynomials are the same degree,divide the coefficients of the highest degree term.
2) If the polynomial in the numerator is a lower degree than the denominator, then y=0 is the asymptote.
3) If the polynomial in the numerator is a higher degree than the denominator, then there is no horizontal asymptote. It is a slant asymptote.
Knowing these three rules, we can determine that there is no horizontal asymptote, since the denominator is a lower degree than the numerator.
Finally, lets find any holes that might be in this graph. Now, just from past knowledge, we should know that no holes will appear in a graph with a slant asymptote. Because of this, lets go ahead and find the slant.
We need to do long division here using both polynomials:
${\displaystyle =\frac{x^2}{x-1}}$
=x−1