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

##### 1 Answer
Feb 23, 2018

See explanation...

#### Explanation:

Alright, So for this question we are looking for six items - holes, vertical asymptotes, horizontal asymptotes, $x$ intercepts, and $y$ intercepts - in the equation $f \left(x\right) = {x}^{2} / \left(x - 1\right)$ 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:
$0 = {x}^{2} / \left(x - 1\right)$

$0 = x$

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:

$= {x}^{2} / \left(x - 1\right)$

$= x - 1$

I'm sorry that there isn't a great way to show you the long divition there, but if you have anymore questions about that, click here.

So there you go, I really hope this helped, and I apologize for the length!
~Chandler Dowd