 ###### 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 3

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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I want to graph a harder example of rational function where the degree of the numerator equals the degree of the denominator. Now, first thing I like to do I find the x intercepts and asymptotes. The x intercepts are much easier to find, if you have your fraction in factored form, which this one is not.

So I’m going to factor this first thing. And the numerator, since we start with the 2x², we’re going to have a 2x and an x, and then the constant terms are going to be probably 1 and 2. I need to get a -3x in the middle and I need to get a -2 here. To get a -2 I need maybe a plus 1, minus 2 or minus 1 plus 2. I think it’s going to be minus 2 plus 1, because first of all that will give me a -2, but second it will given me -4x plus 1 which is -3x. That works.

Now down here, I need an x and an x, and it looks like I’m going to need a 1 and a 4, because 4 minus 1 is 3 I’m going to need maybe a plus 1 minus 4, or minus 1 plus 4, but I think it’s plus 1 minus 4. That will give me minus 4x plus x, -3x. That works.

Now that we have factored form, what are the x intercepts? They come from the zeros of the numerator, and the zeros of this factor is going to be -1/2. We have an x intercept at (-1/2, 0) and the zero of this guy is 2. So (2, 0) is going to be anther x intercept. Let’s plot those. (-½, 0) and 2, 0).

And then we want to look at asymptotes. The zeros of the denominator are -1 and 4, so that gives me x equals -1, x equals 4. And to find the horizontal asymptote, since the degree of the numerator is the same as the degree of the denominator, we look at the leading coefficients of the numerator and denominator. Leading coefficient of the numerator is 2 and of the denominator, 1. So y equals 2/1, y equals 2 is the horizontal asymptote. Let’s graph those.

Y equals 2, is going to go right here and we’ve got x equals -1 and x equals 4, -1 goes right here, 4 goes up here and then we’re ready to plot some points. I usually like to plot points to fill in the regions that the asymptotes and the intercepts divide the plane into. We’ve got this left region here, the region to the left of this asymptote, and we’ve got this middle region and this point’s kind of close to the asymptote, so I’ll probably plot something in between these two intercepts and maybe this point to the right and then I’ll plot something over here. Let’s start with the negative region.

Let’s try -2 to start with, and maybe another point a little to the left of that, afterwards. So -2. the factored form might be easier, 2 times -2 is -4 plus 1 is -3, -2 minus 2 is -4, -2 plus 1, -1 and -2 minus 4 is -6. So we have 12 over 6 and that’s 2. (-2, 2) is a point. And look at this, that point is right on the horizontal asymptote.

Let’s try one to the left. -3; we’ll get 2 times -3, -6, plus 1, -5. -3 minus 2 also -5. -3 plus 1, -2. -3 minus 4, -7. And that gives us 25 over 14. That is about 5/3. So (-3, 5/3), 5/3 is like 1 and 2/3, so it’s about here. It’s very important that you know that even though we did cross the horizontal asymptote, this graph has got to eventually come back and approach the asymptote. And we can plot another point to the left to see if that happens. Maybe we should go a little farther left 2, 3, 4, 5, -5; 2 times -5, -10 plus 1 is -9. -5 minus 2, -7. -5 plus 1, -4, and -5 minus 4, -9. That’s nice, the 9s cancel and I get 7/4. So 7/4 is 1 and 3/4, that’s just a little bit higher than 1 and 2/3. So -5, 1¾, that’s about here. Looks like it’s going to very gently approach this asymptote like so.

Now it crosses through the horizontal asymptote and remember it’s going to have vertical asymptotic behavior too. So it’s got to go up like this. That’s going to be the shape of a left half. Let’s look in the middle region.

I’m interested in what the y intercept is whenever I have a y intercept, so let’s plot zero. That one’s easier. All the x terms are going to be zero, so it’s 1 times -2, -2, over 1 times -4, -4. So that’s a half. (0, 1/2) is here and let’s plot x equals 1. 2 plus 1, 3 times 1 minus 2, -1. 1 plus 1, 2, 1 minus 4, -3. The negatives will cancel and the 3s will cancel and I’ll have again ½, same as here.

So I think what’s going to happen here is I’m going to get like a little lump and it’s gong to come down and approach asymptotically here and here. Now what happens over here? I’m going to plot a point say 5, maybe something else, 5. 2 times 5 plus 1 is 11, 5 minus 2, 3. 5 plus 1, 6, 5 minus 4, 1. Little cancellation helps. I have 11/2 which is about 5.5, actually equals 5.5. So I have 5 and 5.5 3, 4, 5, point 5, it’s way up here. One more point.

Let’s plot, let's say 6 or 7, 7. 2 times 7 plus 1, 15, 7 minus 2, 5. 7 plus 1 is 8 and 7 minus 4 is 3. We get some nice cancellation here and here, so we have 25/8, which is just a hair above 3. So at 7, we’re just a hair above 3, like so. That makes it pretty clear. We’re coming down and approaching the horizontal asymptote like this. Alright and that’s our graph.

This is a little complicated here, but when it passes through the horizontal asymptote, you just have to remember it will come back and approach the asymptote eventually, but it can cross the asymptote in the mean time. Remember your vertical asymptotic behavior, your intercepts, plot all of those if you can and that’s it, that’s your graph.