I’m graphing rational functions, with the degree of the numerators the same as the degree of the denominator. Here’s an example. We have a quadratic over a quadratic. The first thing I want to do is find the x intercepts and the asymptotes. The x intercepts come from the zeros of the numerator. The zero of this guy is 1, so that’s going to give me the x intercept of (1, 0), and the zero of this factor is 3, so that will give me (3, 0). Let me just plot those really quickly, (1, 0) and (3, 0).
And then the asymptotes. Now first of all, let’s note that we’ll get a vertical asymptote if x equals -1 and that appears to be the only vertical asymptote. And then horizontal asymptotes, remember, when the degree of the numerator is the same as the degree of the denominator, we look at the leading coefficient of the numerator and denominator. Now, I don’t really have the leading coefficient laid out for me explicitly, so I'm going to just multiply to get the first term. Now if I multiply the numerator I get an x² and then some stuff. And if I multiply out the denominator, I get an x² and some stuff. Doesn’t mater what the other stuff is, what the other terms are, these are the leading terms. I take the fraction of the leading coefficients and that’s going to be my horizontal asymptote, 1 over 1 which is 1. Let me graph those now.
I have a horizontal asymptote y equals 1, and a vertical asymptote, x equals -1. So x equals -1 goes right here. And y equals 1 goes right here. That takes care of the asymptote. And now, what I want to do is, plot some points before I fill in. I have these two intercepts and I’ve got these asymptote, and that effectively divide the plane up into regions. I’m kind of interested in what happens between these two intercepts, to the right of the intercepts, between this intercept and the asymptote, and then to the left of the asymptote. I’m going to have to plot some points to get the idea of the shape here.
Let’s start with this region to the left. We’re on the negative region, I want to plot some points like let’s say, x equals -5. So I plug in -5 and I get -5 minus 1, -6, -5 minus 3, -8, over -5 plus 1, -4². And that’s 48 over 16, which is 3, so (-5, 3) and I can plot that. -1, 2, 3, 4, 5, 1, 2, 3. That’s about here.
Let’s go a little bit closer to my vertical asymptotes; -3. -3 minus 1, -4, -3 minus 3 is -6, -3 plus 1 is -2, squared is 4. And so the 4s cancel I have minus 1 times by 6, 6. So I have (-3, 6). This is -1, 2, 3, then 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, about here. Now we judge based on the two asymptotes, then I’m going to have a curve, something like this.
Now let’s see what happens in here. I want to plot, x equals zero. If I plug in zero I get -1 times -3, over 1², over 1, so that’s 3. So (0, 3) this point and then, something between the two intercepts. How about 2? 2 minus 1 is 1 times 2 minus 3 is -1, over 2 plus 1, 3² is 9, -1/9. At x equals 2, it's -1/9, just a little bit below zero. That’s sort of interesting. And then let’s plot, say one point to the right of this intercept. 4 or 5 will do. Let’s do 4.
4 minus 1 is 3, times 4 minus 3, 1, over 4 plus 1, 5², 25. That’s 3 over 25, that’s just a hair over zero. Looks like what we’re going to get is, the graph coming in from positive infinity through this point, crossing the horizontal asymptote and then dipping below just a little bit. Let me draw that in, just dipping below a little bit and coming back up and remember it’s got to approach this asymptote as it goes off to infinity. Coming in from the top, something like this. Something like that.
Now remember a graph can't cross its horizontal asymptote. A horizontal asymptote is just a line that the graph approaches as x goes to infinity or as x goes to negative infinity and that does happen here. It’s the vertical asymptotes that can’t be crossed, and the reason they can’t be crossed is, our rational function is not defined at x equals 1.
Remember first find x intercepts and asymptotes, then plot some points, then draw your curve. Remember that asymptotic behavior in the end.