Question

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

Answer 1

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(x)=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:
`0=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:

`=x^2/(x-1)`

`=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