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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Introduction to Rational Functions - 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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I want to talk about vertical asymptotes of rational functions and I ask the question is zero in the denominator enough and to find out I’m going to compare two functions side by side. Y equals 1 over x and y equals x over x times x plus 1.

Now as you’ll notice both of these have x equals zero as a zero of the denominator. So I want to see if that creates a vertical asymptote at x equals zero for both of these guys. Let’s check this one first.

I plug in x and I get y equals 1 over x, 1 over 1 which is 1. I plug in ½ I get 1 over ½ which is 2. I plug in 1/10, 1 over 1/10 is 10, 1 over 100 is 100 and you can see that as these are going to zero these are going to infinity and that means that x equals zero is a vertical asymptote, right? What marks a vertical asymptote is as you get close to that x value your y values go to infinity or negative infinity so it is a vertical asymptote for this guy.

Let’s look at this function now. When I plug in 1 I get 1 over 1 times, 1 plus 1, 2, ½. I plug in ½, I get ½ over ½ times ½ plus 1, 3/2. I get cancellation. 1 over 3/2 is 2/3. Plug in 1/10; I get 1 over 10 divided by 1/10 and 1/10 plus 1, 11/10. These cancel and I get the reciprocal of 11/10, 10/11. You can see that this is not actually going to infinity. 1/100, one more, 1/100 times 1/100 plus 1 that’s 101/100. These cancel, you get 100/101. Doesn’t look like it’s going to infinity. It does kind of look like it’s headed towards the value 1 but as this goes to zero this is not going to infinity. So x equals zero is not an asymptote.

Now what do we learn from this? The bottom line is if you look at a rational function and you can cancel the factor that gives you zero in the denominator, you’re not going to get a vertical asymptote there and that’s the key the fact that these xs would cancel. If I cancel them I get 1 over x plus 1 and then I make the note x not equal to zero.

This function is actually 1 over x plus 1 which is your translated y equals 1 over x graph with a little hole at x equals zero. It doesn’t have an asymptote there.

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