# Introduction to Rational Functions - Problem 1

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.

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