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(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