 ###### 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.

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# Graphing Rational Functions, n>m - 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.

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We're graphing rational functions where the degree of the numerator is bigger than the degree of the denominator. Here’s an example; y equals x times quantity x plus 4 over 4 minus x. And the first thing I like to do is find the x intercepts and the asymptotes.

The x intercepts come from the zeros of the numerator and the numerator has zeros at zero and -4. So we get 0,0 and -4 ,0 and just plot those. 0,0 and -4,0. Now I want to find the asymptotes.

First, observe that x equals 4 is going to be a vertical asymptote because it makes the denominator 0. And for the non-vertical asymptote, we can tell that there’s not going to be a horizontal asymptote because the numerator has degree bigger than the denominator. So we have to use polynomial division to find if there’s no oblique asymptote.

Let me expand this numerator x² plus 4x. So I have to divide x² plus 4x by 4 minus x which is the same as -x plus 4. So let me do that. -x plus 4 divided by x² plus 4x. So we'll have to multiply by -x to get x², -x. I multiply through and I get x² minus 4x, I change the signs and I add. And this gives me 8x and what do I multiply by -x to get 8x, -8. And I multiply through and I get 8x minus 32. I change the signs and I add and I get a remainder of 32.

So this function can actually be written as, y equals the quotient -x minus 8 plus 32, over -x plus 4. Now if you look at this you can see that as x gets very large, this term is going to go to zero. The denominator is going to get pretty large then negative direction. This term will go to zero and the graph will be more and more like the line y equals -x minus 8. That’s the oblique asymptote. So I’m going to graph that now. -x minus 8 has a slope of -1 and it passes through -8. So I plot two points then graph it. And this is our oblique asymptote; y equals -x minus 8.

We also have a vertical asymptote of x equals 4, so let me draw that, x equals 4. Here we go, x equals 4. And so now what I need to do, is plot points to see where the graph goes. I can already kind of tell what the graph is going to do in here. It looks like it’s going to go down below the x axis in here and go up and approach these two asymptotes. Just to be sure though, I may plot a point between -4 and 0 to see what happens and so let me do that.

Let’s plot -2. When x equals -2 let’s look up here. I think this will be easiest way to calculate, -2 times -2 plus 4, 2. There’s -4 on top, 4 minus -2 is 4 plus 2 6. So -2/6. So we get -2 -2/3 and this is -4 -2 this is going to be a little bit below the x axis. So I think it does actually go something like this. And then over and approaching this oblique asymptote like that.

Let’s see what happens over here to the right of the vertical asymptote. Let’s plot x equals 8. I get 8 times 12, don’t be in too much of a hurry to multiply because you’ll make a cancellation. Over 4 minus 8 -4 and the -4 does cancel with the 8 leaving 2. So we have 2 times 12, 24, over -1, -24. (8,-24) that’s down here.

Let me plot another point let’s say 12. I get 12 times 16 over 4 minus 12, -8. So -8 and 16 cancel leaving a 2 and again I get 24 over -1, -24. That’s interesting. So 12 I get -24. And I think I’m going to get the same basic shape I get up here this kind of downward opening 'u' shape.

So we’ve got our oblique asymptote; y equals -x minus 8 and we’ve got our vertical asymptote. And we’ve got to graph it looks actually like it's symmetric about this point right here. So this is our graph.