##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics

# Graphing Rational Functions, n=m - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I want to graph some rational functions, where the degree of the numerator is equal to the degree of the denominator. Here’s an example; I have a rational function where the numerator as a quadratic and the denominator is also a quadratic so we have a degree 2 over a degree 2. First thing I want to do is, identify the x intercepts and the asymptotes.

The x intercepts come from the zeros in the numerator and since the numerator is factored, it’s very easy to see the two and 4s are zeros, and that gives me x intercepts of (2, 0) and (4, 0), and I’m going to plot those immediately, (2, 0) and (4, 0).

The the vertical asymptotes come from the zeros, the denominator, x equals zero and x equals 3. The horizontal asymptote comes from the leading coefficients of the numerator and denominator. So you could just mentally multiply out the numerator just far enough to know what the leading term is, x times x, x², and there’s other terms, x times x, again x² plus some other terms, and the leading coefficients are 1 and 1. So the horizontal asymptote, is going to be y equals 1/1, y equals 1.

Let me plot those asymptotes while they’re fresh in my mind. We’ve got x equals zero, that’s the y axis. We’ve got x equals 3, right between our 2 intercepts and we’ve got y equals 1, right here. Now what I want to do is plot some points, to kind of flash out what happens in between the asymptotes and the intercepts. Let’s start with this left hand region. Let’s plot some points in the negative part of the plane.

Let’s try -1. If I plug in -1 I get -3, times -5, that’s going to be 15 over -1 times -4, that’s 4, 15 over 4 that’s 3 and ¾. So (-1, 3¾) that’s about here. Let’s try something a little further to the left like -3. -3 minus 2, -5, times -3 minus 4, -7. -3 times -6, so we get 35/18. That’s not a great number but it’s very, very close to 36/18 which is 2. So I’ll plot something like -3,2 somewhere like here, a little bit less than 2 but close. And that’s actually enough information to draw this branch of our function. It’s going to go up this asymptote, and down to approach this one. Something like that.

Now let’s focus on the region between these two vertical asymptotes. We’ve got this intercept, it’s going to pass through the axis here but I want to know whether it’s below the x axis or above the x axis at this point, so let me plot x equals 1. I get 1 minus 2, -1, 1 minus 4, -3, and 1 minus 3, -2. So I have 3 over -2, -3/2. At 1 I have -3/2 at about here. Now what’s probably going to happen is, the graph is going to go down and approach this vertical asymptote. It’s going to have asymptotic behavior in this direction, and it has to have asymptotic behavior in this direction as well. So it’s going to have to go up like this, through the horizontal asymptote and just go up and approach the asymptote from the left something like that. Let’s graph this part.

What I want to do is figure out what happens to the right of this intercept, let’s plot, 1, 2, 3, , 4, 5, x equals 5. I get 5 minus 2, 3 times 5 minus 4, 1. 5 minus 3, 5 times 2. So this is 3 over 10, .3. It’s about a third. So at this point I am at 5.3 about here. I am going to approach this line asymptotically and this one asymptotically. That’s what my graph looks like. I’ve got a branch up here, I've got a piece that crosses the horizontal asymptote in the middle, and then this piece goes against this vertical asymptote and up, and approaches this horizontal asymptote from below.

Remember we can cross horizontal asymptotes. A graph can cross horizontal asymptotes a number of times actually, but in the end they have to approach them to the right and to the left. It’s the vertical asymptotes that can never be crossed because the function’s not defined at those points; in this case zero and 3.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete