# Introduction to Rational Functions - Problem 3

###### Transcript

Let’s take a look at these functions and find out what our vertical asymptotes are. Remember that in order to identify the vertical asymptotes, you really have to factor both the numerator and denominator just to see if there are any common factors and in the numerator I get 2 times x plus 3 in the denominator I get x, x, I’m going to have, well I need maybe a 3 and a 5, x minus 5, x plus 3, does that work? Minus 5x plus 3x is -2x, that works.

And then you notice that the x plus 3s cancel. You can cancel those but you have to make sure that you make a little note 'x not equal to -3. Now why do you have to do that? Well, in this function if you plug in -3, it’s undefined. In this function you have to also make it undefined. It has to have the same domain to be the same function, so this domain restriction is absolutely necessary. What it tells you is that x equals -3 is not a vertical asymptote, the vertical asymptote is going to be x equals 5. That’s your asymptote.

Let’s take a look at this one. Again we factor the numerator and denominator and we get, I’m going to need an x and an some factors of a, to get 7 I think we need 8 and 1,let's use plus 8 and minus 1. That will give us a -8 here, so I’ll have 8x minus x that will give us 7x. And then we have 7 minus 1 in the denominator. Again we can cancel these provided we make the note this is x plus 8 with x not equal to 1. This function has no vertical asymptotes. It’s never undefined, this part is never undefined but it’s not defined for 1. There are no vertical asymptotes here.

Now in case you’re wondering these domain restrictions they do affect the graph. These are going to create little holes in the graph. In this case you have the graph of a line with a hole at x equals 1. In this case you’ve got a transformed 1/x graph with a hole at x equals -3. We’ll get to graphing functions with holes later.

Let’s take a look at this one. Factor the numerator and denominator and I get x times x² minus 4 over and this is, this is actually a perfect square right. X minus 2 quantity squared. Now I can factor this difference of squares. It's x minus 2, x plus 2 and the denominator is x minus 2 times x minus 2. You can cancel these and notice you don’t even have to make the note. You don’t have to make the note that x doesn’t equal 2 because you still have x minus 2 in the denominator. There is no reason to write x doesn’t equal 2 if it can, still can equal 2. You’ll notice that even though we cancelled an x minus 2, x equals 2 is still an asymptote. If you still have a factor in the denominator, that creates the asymptotes, you’re going to have it, x equals 2 is an asymptote.

Now let’s take a look at this guy. Now this numerator doesn’t factor but the denominator is x minus 3², there’s really nothing to cancel here and so we would say x equals 3 is our vertical asymptote, that’s the zero in the denominator.

The bottom line is you do have to look at the zeros in the denominator but you have to make sure that any factors that create zero in the denominator don’t cancel. If they do cancel you should cancel them, rewrite the function and look at the reduced rational function to decide what your vertical asymptotes are.