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 less than m - Problem 2

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, I want to try a slightly harder example; graph y equals 36 minus 9x over the quantity x minus 6². Usually the first thing I do, is plot the x intercepts and the asymptotes. The x intercepts come from the zeros of the numerator, and this numerator is going to be zero whenever 9x equals 36, so when x equals 4. So my x intercept is going to be (4, 0). I can plot that now.

Now, as far as the asymptotes go, the asymptotes are going to come from zeros of the denominator, provided that they aren’t also zeros on the numerator, so we have to check that, but here the zeros of the denominator are x equals 6, and the zero of the numerator is 4. So x equals 6 is going to be a vertical asymptote, and I can draw that in. x equals 6, right here. And because the denominator has a degree bigger than the numerator, y equals zero is going to be the horizontal asymptote. So y equals zero, the x axis will be a horizontal asymptote for my graph, I just have to remember that.

And what I need to do now, is plot some points to sort of figure out what happens everywhere else. I think a good point to plot would be x equals zero. We want to know what the y intercept’s going to be. So when x equals zero, I get 36 over -6². That’s going to be 36/36, which is 1. So (0, 1) is right here. Let's plot 2. We get 36 minus 18, over 2 minus 6, -4², which is 16. So 18 over 16 is 9/8, so 1 and 1/8, just a little more than 1, so here. Keeping in mind that y equals zero is an asymptote and that this point’s a little higher than this point, it looks like it’s going to come up, and then very slowly go down toward this asymptote.

Now what do you think it does as x goes closer to 6? I’m guessing it goes down to negative infinity. We could plot a point to find, but I’m pretty sure. Let’s draw this part of the graph. Now what happens on the other side? I need to plot a point say at 8, to find out. Now at x equals 8 we get 36 minus 9 times 8 which is 72. 36 minus 72 is -36 and 8 minus 6 is 2, 2² is 4. I get -9. -9 is way down here, so 8, it’s down here. So looks like it’s going to be negative.

Now let’s see what happens let’s say, so this is 10, 12, it’s way down there, so I’m going to plug in 12 to see how far it’s come up. I’m assuming it’s going to come up, so it can have y equals zero as an asymptote. When I plug in 12, I get 36 minus 9 times 12, that’s 36 minus 108. And in the denominator I get 12 minus 6, 6², so 36. Now 36 minus 108 is -72, over 36, -2. It’s come up quite a bit. At 12, it’s -2. I think it’s going to be something like this.

Now again, the things you have to remember, the first thing when you’re graphing a rational function is, plot the x intercepts and asymptotes. You know when the degree of the numerator is less than the degree of the denominator, y equals zero is going to be a horizontal asymptote. Make sure that your graph in the end has that asymptotic behavior, that comes down and approaches y equals zero.

A lot of people will say you can’t cross an asymptote, you actually can and we do cross the horizontal asymptote right here. It’s the vertical asymptotes you can’t cross, because a vertical asymptote represents a point, a number 6, where the function’s undefined. You get division by zero at x equals 6, and that’s why you can’t cross the vertical asymptote. But horizontal asymptote gets crossed all the time so, don’t get confused about that. Then once you’ve got the asymptotes and your intercepts plotted, just plot a few points to get an idea of the shape, and you’ll get a pretty good graph almost every time.