##### 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
- Attend and watch FREE live webinar on useful 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 another rational function where the degree of the numerator is bigger than the degree of the denominator. Now here, usually the first thing I like to do, is find the x intercepts and the asymptotes. But I’m going to need the numerator to be factored in order to that, so let’s factor the numerator.

We’ve got an x² here, so the factors are going to start with and x and we have a 3 here. So I’m hoping that a 1 and a 3 will work. It looks like it will because 3 minus 1 is 2, so let’s try -3 and plus 1. That will definitely give me my -3 and it will give me -3x plus x which is -2, so it does work. And so that’s my factored form. And remember that x intercepts come for the zeros of the numerator, so I have an intercept at -1,0 and at 3,0 and lets plot those right away. -1,0 is right here and 3,0 is over here.

Now let’s find the asymptotes. Where you get a vertical asymptote when x equals 0 because you got the rational by 0, and the non-vertical asymptote, the oblique asymptote if it exists will find it by polynomial division. Remember that a rational function does not have a horizontal asymptote if the numerator has a higher degree than the denominator. So we have to look for an oblique asymptote instead.

So we get x² minus 2x minus 3 divided by 2x. So what do we need to multiply by 2x to get x²? 1/2x. I multiply through and I get x² and then I subtract, -2x. What do I multiply by 2x to get -2x? -1. Multiply 2 to get -2x change signs and add and my remainder is going to be -3.

So what this means is, that my function can actually be written y equals this thing, the quotient, 1/2x minus 1 plus the remainder of -3 over the divisor 2x.

Now if you look at this part of the function, you can see that as x gets very large, this thing is going to diminish away to 0 and you’ll be left with this line. That means that the graph is going to approach this line of slope 1/2 and y intercept -1. So that’s going to be our oblique asymptote. And we’ll plot that; y equals 1/2 x minus 1. Well minus 1 is here. A slope of 1/2 but it's going to go up 1 over 2, so it will pass through here. So let me draw that line. It looks something like that.

That’s y equals 1/2 x minus 1. And don’t forget we also have our vertical asymptote x equals 0 and that’s just the y axis. So the vertical asymptote in the intercepts divide the plane up into regions, let’s plot some points. How about -2?

So what happens when x is -2? If I look up here, I get -2 plus 1, -1, -2 minus 3 -5 so -1 times -5 over 2 times -2, -4. So it's 5 over -4, -5/4. (-2 -5/4) that’s just a quarter less than -1, so it will be down here. That’s actually enough that we can probably graph the left half. Remember that we have a vertical asymptote here, so it's going to have to go through and up approach the y axis asymptotically, and then down and approach the oblique asymptote.

What happens over here? Let’s plot a point at x equals 1, see what happens. X equals 1 we get 1 plus 1, 2 and 1 minus 3, -2. And the denominator we get 2 times 1, 2 the 2’s cancel we get -2 so (1,-2).

And actually put it a little bit lower. That might be enough to finish the graph. It's going to approach this asymptote and it will have to go down and approach the vertical asymptote, so something like this. And that’s all there is to it.

Just find your intercepts and your asymptotes and then plot a few points just to flash out your graph.

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!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### 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