##### 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

# Introduction to Rational Functions - Problem 1

###### 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 talk about functions that are related to the function y equals 1/x. Now notice the function 1/x is a rational function, rational functions are polynomials divided by other polynomials and here one is a polynomial, it’s a constant function which is a polynomial and x is a polynomial.

We’re pretty familiar with this it’s one of our parent functions. We’re familiar with the graph it’s got a horizontal asymptote and a vertical asymptote but what’s really interesting is that functions of this form, f of x equals any linear function divided by another linear function are transformations of y equals 1/x. Now let’s see that in an example. Y equals -x plus 1 over x plus 3.

The first thing I want to do is write this in another form and to get it in another form I’m going to divide -x plus 1 by x plus 3, and I get -1 and I multiply through -x minus 3 and then I change signs and add and so I’ll get 4 as a remainder and that means that this function can also be written as negative1, that’s my quotient, plus the remainder, 4, divided by the divisor. 4 over x plus 3.

And when you have it in this form it’s really easy to see what transformation you have. First you’ve got a horizontal shift because of this x plus 3, you’ve got a vertical stretch because of the 4 and a vertical shift. So this is my function, x, -1 plus 4 over x plus 3. Now we can get rid of this. I want to use transformations to graph this functions so let me start with u and 1 over u. Now till you make the connection between these two I’m going to set u equals to x plus 3. U equals x plus 3, so x equals u minus 3. And then some basic points that we’ll need, minus 1, minus 1, 1, 1. We’ll also need to use the face that we’ve got two asymptotes. X equals zero and y equals zero are our asymptotes.

What happens to these points as we transform them? Well based on this transformation x equals us minus 3, I take the u value and subtract 3 so I get -4, -2. And I can also apply that transformation to the vertical asymptote because it’s an x value. Right? So and actually think of this as u equals zero. So x equals u minus 3, you subtract 3 and you get x equals -3. I can draw that asymptote right now. U equals -3 because it’s right here. And what are the y values going to be?

Well notice that we’re taking our 1 over u and remember that u s x plus 3 and we’re multiplying it by 4 and subtracting by 1. So take these values, multiply by 4 and subtract 1. We get -4 minus 1, -5. Multiply by 4 we get 4, subtract 1 we get 3. That gives us two points, -4 and -5, which is -4 is here, which is 1, 2, 3, 4, 5, about there, oops, a little further down. And then (-2, 3), that’s up here. Looks like we might need a few more points so let’s use -2,-½ and 2,½.

For the transformations, for the horizontal transformations all we have to do is subtract 3 so we get -5 here and here we get -1. And again for the y value we multiply by 4 and subtract 1. So multiply this by 4 and you get -2, minus 1, -3. Multiply by 4 we get 2, minus 1 is 1. So -5,-3. So this was -4, -5 and -3 is here and then -1,1. That’s here. So now I have some idea on what this graph’s going to look like.

Wait a second, I need to do one more thing, sorry. The horizontal asymptote that’s going to transform too. It’s a y value so I have to apply the vertical transformation times 4 minus 1. Y equals zero. Zero times 4 is zero minus 1 is, y equals -1. That’s very important. I need to know where the horizontal asymptote ends up so I can get the correct asymptotic behavior in my graph. There’s my horizontal asymptote, y equals -1.

Now I can finish my graph. So we’ve got something like this and something like this. It’s really just as easy as that to graph the transformation of 1 over x. Just remember whenever you have a function of this type, a linear function divided by another linear function, just as long as they don’t both have the same factor they’re going to give you a graph that looks like this, some kind of transformation of y equals 1 over x.

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