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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.

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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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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